Examkin

Trigonometric Graphs and Simple Equations

24 questions

All types24Multiple choice10Numeric14
Question 1Question →

For the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ, solve the trigonometric equation 2cos⁡2x−cos⁡x−1=02\cos^2 x - \cos x - 1 = 0. Which set contains all the solutions for xx?

Show answer & explanation

Answer: 0∘,120∘,240∘,360∘0^\circ, 120^\circ, 240^\circ, 360^\circ

Answer

The complete set of solutions is 0∘,120∘,240∘,360∘0^\circ, 120^\circ, 240^\circ, 360^\circ.
Factoring 2cos⁡2x−cos⁡x−1=02\cos^2 x - \cos x - 1 = 0 gives (2cos⁡x+1)(cos⁡x−1)=0(2\cos x + 1)(\cos x - 1) = 0. Setting the first factor to zero yields cos⁡x=−12\cos x = -\frac{1}{2}, which has solutions at 120∘120^\circ and 240∘240^\circ in the interval [0∘,360∘][0^\circ, 360^\circ]. Setting the second factor to zero gives cos⁡x=1\cos x = 1, which has solutions at 0∘0^\circ and 360∘360^\circ. Combining these yields the set 0∘,120∘,240∘,360∘0^\circ, 120^\circ, 240^\circ, 360^\circ.

Step-by-Step Solution

1
Factor the quadratic trigonometric equation
(2cos⁡x+1)(cos⁡x−1)=0(2\cos x + 1)(\cos x - 1) = 0
Treat cos⁡x\cos x as a single variable to simplify into standard quadratic factors.
2
Set each factor to zero to find values for cos⁡x\cos x
cos⁡x=1\cos x = 1 or cos⁡x=−12\cos x = -\frac{1}{2}
Zero-product property requires at least one factor to be zero.
3
Solve for xx in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ
For cos⁡x=1\cos x = 1: x=0∘,360∘x = 0^\circ, 360^\circ. For cos⁡x=−12\cos x = -\frac{1}{2}: x=180∘−60∘=120∘x = 180^\circ - 60^\circ = 120^\circ (Quadrant II) and x=180∘+60∘=240∘x = 180^\circ + 60^\circ = 240^\circ (Quadrant III).
Cosine is negative in Quadrants II and III, with a reference angle of 60∘60^\circ.
4
Combine all unique solutions in ascending order
x=0∘,120∘,240∘,360∘x = 0^\circ, 120^\circ, 240^\circ, 360^\circ
Include all solutions within the given domain boundaries.

Key Concept

Solving Quadratic Trigonometric Equations
Question 2Question →

Find the smallest positive value of θ\theta (in degrees) satisfying the trigonometric equation sin⁡(3θ)+3cos⁡(3θ)=2\sin(3\theta) + \sqrt{3}\cos(3\theta) = \sqrt{2}.

Show answer & explanation

Answer: 25

Answer

The smallest positive value of θ\theta is 25∘25^\circ.
Dividing the equation by 22 reduces it to sin⁡(3θ+60∘)=22\sin(3\theta + 60^\circ) = \frac{\sqrt{2}}{2}. The principal acute angle is 45∘45^\circ. The second quadrant angle giving a positive sine is 180∘−45∘=135∘180^\circ - 45^\circ = 135^\circ. Setting 3θ+60∘=135∘3\theta + 60^\circ = 135^\circ gives 3θ=75∘3\theta = 75^\circ, which yields θ=25∘\theta = 25^\circ. This is smaller than any positive solution generated by the first quadrant branch.

Step-by-Step Solution

1
Transform the left-hand side into a single harmonic function Rsin⁡(3θ+α)R\sin(3\theta + \alpha).
Dividing the equation by 22 yields 12sin⁡(3θ)+32cos⁡(3θ)=22\frac{1}{2}\sin(3\theta) + \frac{\sqrt{3}}{2}\cos(3\theta) = \frac{\sqrt{2}}{2}, which simplifies to sin⁡(3θ+60∘)=22\sin(3\theta + 60^\circ) = \frac{\sqrt{2}}{2}.
The identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B allows us to combine sin⁡(3θ)\sin(3\theta) and cos⁡(3θ)\cos(3\theta) using cos⁡60∘=12\cos 60^\circ = \frac{1}{2} and sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}.
2
Find the quadrant solutions for the angle 3θ+60∘3\theta + 60^\circ.
First quadrant: 3θ+60∘=45∘+360∘k  ⟹  3θ=−15∘+360∘k3\theta + 60^\circ = 45^\circ + 360^\circ k \implies 3\theta = -15^\circ + 360^\circ k.
Second quadrant: 3θ+60∘=135∘+360∘k  ⟹  3θ=75∘+360∘k3\theta + 60^\circ = 135^\circ + 360^\circ k \implies 3\theta = 75^\circ + 360^\circ k.
Since the sine of the angle is positive (22\frac{\sqrt{2}}{2}), solutions exist in both the first (45∘45^\circ) and second (180∘−45∘=135∘180^\circ - 45^\circ = 135^\circ) quadrants.
3
Calculate the smallest positive angle θ\theta.
For k=0k = 0 in the second quadrant branch, 3θ=75∘  ⟹  θ=25∘3\theta = 75^\circ \implies \theta = 25^\circ. (The first quadrant branch yields θ=−5∘\theta = -5^\circ for k=0k=0 and θ=115∘\theta = 115^\circ for k=1k=1). Thus, θ=25∘\theta = 25^\circ is the smallest positive solution.
Comparing all non-negative resulting angles demonstrates that 25∘25^\circ is the smallest strictly positive solution.

Key Concept

Solving linear trigonometric equations of the form Asin⁡x+Bcos⁡x=CA\sin x + B\cos x = C using RR-formula reduction.
Question 3Question →

Find the sum, in degrees, of all solutions to the trigonometric equation 3tan⁡(2x)=3\sqrt{3}\tan(2x) = 3 in the interval 0∘≤x≤180∘0^\circ \le x \le 180^\circ.

Show answer & explanation

Answer: 150

Answer

The sum of all solutions to the equation in the given interval is 150 degrees.
Isolating tan⁡(2x)\tan(2x) gives 3\sqrt{3}. For 0∘≤2x≤360∘0^\circ \le 2x \le 360^\circ, tan⁡(2x)=3\tan(2x) = \sqrt{3} yields solutions at 2x=60∘2x = 60^\circ and 2x=240∘2x = 240^\circ. Dividing by 2 gives x=30∘x = 30^\circ and x=120∘x = 120^\circ. Adding these solutions yields 30∘+120∘=150∘30^\circ + 120^\circ = 150^\circ.

Step-by-Step Solution

1
Isolate the trigonometric function
tan⁡(2x)=33=3\tan(2x) = \frac{3}{\sqrt{3}} = \sqrt{3}
Dividing both sides by \sqrt{3} simplifies the expression to a standard special angle ratio.
2
Determine the domain for the argument 2x2x
Since 0∘≤x≤180∘0^\circ \le x \le 180^\circ, multiplying the inequality by 2 gives 0∘≤2x≤360∘0^\circ \le 2x \le 360^\circ.
This establishes the range of angles to search for 2x2x within one complete turn.
3
Find all values of 2x2x where tangent equals 3\sqrt{3}
2x=60∘2x = 60^\circ (1st quadrant) and 2x=180∘+60∘=240∘2x = 180^\circ + 60^\circ = 240^\circ (3rd quadrant)
The tangent function is positive in Quadrants I and III with a reference angle of 60∘60^\circ.
4
Solve for xx
x=60∘2=30∘x = \frac{60^\circ}{2} = 30^\circ and x=240∘2=120∘x = \frac{240^\circ}{2} = 120^\circ
Dividing each angle by 2 yields the values of xx lying within the domain 0∘≤x≤180∘0^\circ \le x \le 180^\circ.
5
Calculate the sum of the solutions
30∘+120∘=150∘30^\circ + 120^\circ = 150^\circ
The question specifically requests the sum of all valid solutions.

Key Concept

Solving trigonometric equations using reference angles and domain transformation
Estimated Time:2m 0s
Question 4Question →

Find the total number of distinct solutions to the trigonometric equation 2cos⁡2(2x)+sin⁡(2x)−1=02\cos^2(2x) + \sin(2x) - 1 = 0 in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ.

Show answer & explanation

Answer: 6

Answer

The total number of distinct solutions is 6.
Substituting cos⁡2(2x)=1−sin⁡2(2x)\cos^2(2x) = 1 - \sin^2(2x) gives the quadratic 2sin⁡2(2x)−sin⁡(2x)−1=02\sin^2(2x) - \sin(2x) - 1 = 0, which factors into (2sin⁡(2x)+1)(sin⁡(2x)−1)=0(2\sin(2x) + 1)(\sin(2x) - 1) = 0. For 0∘≤x≤360∘0^\circ \le x \le 360^\circ, the angle argument 2x2x covers 0∘≤2x≤720∘0^\circ \le 2x \le 720^\circ. The equation sin⁡(2x)=1\sin(2x) = 1 provides 2 values for xx (45∘,225∘45^\circ, 225^\circ), while sin⁡(2x)=−12\sin(2x) = -\frac{1}{2} provides 4 values for xx (105∘,165∘,285∘,345∘105^\circ, 165^\circ, 285^\circ, 345^\circ). Summing these gives 6 distinct solutions in total.

Step-by-Step Solution

1
Use the Pythagorean trigonometric identity cos⁡2(2x)=1−sin⁡2(2x)\cos^2(2x) = 1 - \sin^2(2x) to express the entire equation in terms of sin⁡(2x)\sin(2x).
2(1−sin⁡2(2x))+sin⁡(2x)−1=0  ⟹  2−2sin⁡2(2x)+sin⁡(2x)−1=02(1 - \sin^2(2x)) + \sin(2x) - 1 = 0 \implies 2 - 2\sin^2(2x) + \sin(2x) - 1 = 0
Converting all trigonometric terms to a single function allows the equation to be solved as a polynomial.
2
Rearrange and factorize the resulting quadratic equation in terms of sin⁡(2x)\sin(2x).
2sin⁡2(2x)−sin⁡(2x)−1=0  ⟹  (2sin⁡(2x)+1)(sin⁡(2x)−1)=02\sin^2(2x) - \sin(2x) - 1 = 0 \implies (2\sin(2x) + 1)(\sin(2x) - 1) = 0
Factorization splits the quadratic trigonometric equation into two simple linear trigonometric equations.
3
Determine the expanded domain for 2x2x given 0∘≤x≤360∘0^\circ \le x \le 360^\circ.
0∘≤2x≤720∘0^\circ \le 2x \le 720^\circ
Multiplying the bounds of xx by 2 accounts for two full rotations in the unit circle.
4
Solve the first linear equation sin⁡(2x)=1\sin(2x) = 1 within 0∘≤2x≤720∘0^\circ \le 2x \le 720^\circ.
2x=90∘,450∘  ⟹  x=45∘,225∘2x = 90^\circ, 450^\circ \implies x = 45^\circ, 225^\circ (2 distinct solutions)
The sine function equals 1 at 90∘90^\circ in the first revolution and at 90∘+360∘=450∘90^\circ + 360^\circ = 450^\circ in the second revolution.
5
Solve the second linear equation sin⁡(2x)=−12\sin(2x) = -\frac{1}{2} within 0∘≤2x≤720∘0^\circ \le 2x \le 720^\circ.
2x=210∘,330∘,570∘,690∘  ⟹  x=105∘,165∘,285∘,345∘2x = 210^\circ, 330^\circ, 570^\circ, 690^\circ \implies x = 105^\circ, 165^\circ, 285^\circ, 345^\circ (4 distinct solutions)
The sine function is negative in the 3rd and 4th quadrants of both revolutions.
6
Combine the solution counts from both cases.
Total number of solutions = 2+4=62 + 4 = 6.
Adding the valid solutions from both factor equations gives the complete set of roots.

Key Concept

Solving quadratic trigonometric equations across multiple revolutions
Estimated Time:3m 0s
Question 5Question →

What is the period of the trigonometric function y=3cos⁡(4x)y = 3\cos(4x)?

Show answer & explanation

Answer: 90∘90^\circ

Answer

The period of the given trigonometric function is 90∘90^\circ.
The period of a function of the form y=acos⁡(bx)y = a\cos(bx) is given by T=360∘bT = \frac{360^\circ}{b}. Substituting b=4b = 4 yields T=360∘4=90∘T = \frac{360^\circ}{4} = 90^\circ.

Step-by-Step Solution

1
Identify the standard form parameters
For y=3cos⁡(4x)y = 3\cos(4x), amplitude a=3a = 3 and coefficient of xx is b=4b = 4.
The general form of a cosine function is y=acos⁡(bx+c)+dy = a\cos(bx + c) + d.
2
Apply the period formula for trigonometric functions in degrees
Period T=360∘b=360∘4=90∘T = \frac{360^\circ}{b} = \frac{360^\circ}{4} = 90^\circ.
The full cycle of a standard cosine graph completes in 360∘360^\circ, so multiplying the input by bb compresses the period by a factor of bb.

Key Concept

Period of a Cosine Function
Estimated Time:45s
Question 6Question →

Which of the following sets contains all the values of xx in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ that satisfy the trigonometric equation 2cos⁡2x+3sin⁡x−3=02\cos^2 x + 3\sin x - 3 = 0?

Show answer & explanation

Answer: 30∘,90∘,150∘30^\circ, 90^\circ, 150^\circ

Answer

The values of xx in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ satisfying the equation are 30∘,90∘,30^\circ, 90^\circ, and 150∘150^\circ.
By substituting cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x, the equation reduces to 2sin⁡2x−3sin⁡x+1=02\sin^2 x - 3\sin x + 1 = 0, which factors into (2sin⁡x−1)(sin⁡x−1)=0(2\sin x - 1)(\sin x - 1) = 0. Solving sin⁡x=1/2\sin x = 1/2 gives x=30∘x = 30^\circ and x=150∘x = 150^\circ within the specified domain. Solving sin⁡x=1\sin x = 1 gives x=90∘x = 90^\circ. Combining these yields the set of solutions 30∘,90∘,150∘30^\circ, 90^\circ, 150^\circ.

Step-by-Step Solution

1
Use the Pythagorean trigonometric identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x to rewrite the equation in terms of sin⁡x\sin x.
2(1−sin⁡2x)+3sin⁡x−3=0  ⟹  2−2sin⁡2x+3sin⁡x−3=02(1 - \sin^2 x) + 3\sin x - 3 = 0 \implies 2 - 2\sin^2 x + 3\sin x - 3 = 0
Converting the equation to involve a single trigonometric function allows it to be solved as a quadratic equation.
2
Simplify and rearrange the equation into standard quadratic form.
−2sin⁡2x+3sin⁡x−1=0  ⟹  2sin⁡2x−3sin⁡x+1=0-2\sin^2 x + 3\sin x - 1 = 0 \implies 2\sin^2 x - 3\sin x + 1 = 0
Multiplying by −1-1 simplifies factoring.
3
Factor the quadratic equation (2sin⁡x−1)(sin⁡x−1)=0(2\sin x - 1)(\sin x - 1) = 0 to solve for sin⁡x\sin x.
sin⁡x=12\sin x = \frac{1}{2} or sin⁡x=1\sin x = 1
Setting each linear factor to zero yields the possible values for sin⁡x\sin x.
4
Determine all values of xx in the domain 0∘≤x≤360∘0^\circ \le x \le 360^\circ for each case.
For sin⁡x=12\sin x = \frac{1}{2}, x=30∘x = 30^\circ and x=180∘−30∘=150∘x = 180^\circ - 30^\circ = 150^\circ. For sin⁡x=1\sin x = 1, x=90∘x = 90^\circ.
Sine is positive in the first and second quadrants, and equals 1 at 90∘90^\circ.

Key Concept

Solving quadratic trigonometric equations by using fundamental identities to express the equation in terms of a single trigonometric function.
Estimated Time:2m 0s
Question 7Question →

For the domain 0∘≤x≤360∘0^\circ \le x \le 360^\circ, solve the trigonometric equation 2sin⁡x+3=02\sin x + \sqrt{3} = 0. Which of the following gives the complete set of values for xx?

Show answer & explanation

Answer: 240∘240^\circ and 300∘300^\circ

Answer

x=240∘x = 240^\circ and x=300∘x = 300^\circ
Rearranging 2sin⁡x+3=02\sin x + \sqrt{3} = 0 gives sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}. The reference angle for which sine equals 32\frac{\sqrt{3}}{2} is 60∘60^\circ. Since sine is negative in the third and fourth quadrants, the solutions are 180∘+60∘=240∘180^\circ + 60^\circ = 240^\circ and 360∘−60∘=300∘360^\circ - 60^\circ = 300^\circ.

Step-by-Step Solution

1
Isolate the trigonometric function in the equation
sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}
Subtract 3\sqrt{3} from both sides and divide by 22.
2
Find the basic reference angle
Reference angle α=60∘\alpha = 60^\circ
sin⁡(60∘)=32\sin(60^\circ) = \frac{\sqrt{3}}{2}.
3
Determine the relevant quadrants
Quadrants III and IV
The sine function is negative in the third and fourth quadrants.
4
Calculate the solutions within the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ
Quadrant III: x=180∘+60∘=240∘x = 180^\circ + 60^\circ = 240^\circ; Quadrant IV: x=360∘−60∘=300∘x = 360^\circ - 60^\circ = 300^\circ
Apply quadrant reduction formulas for Quadrant III (180∘+α180^\circ + \alpha) and Quadrant IV (360∘−α360^\circ - \alpha).

Key Concept

Solving simple trigonometric equations using reference angles and quadrant rules
Question 8Question →

A sine function is given by the equation y=5sin⁡(23x)−2y = 5\sin\left(\frac{2}{3}x\right) - 2. What is the period of this trigonometric function in degrees?

Show answer & explanation

Answer: 540

Answer

The period of the trigonometric function is 540∘540^\circ.
For a general sine curve of the form y=asin⁡(bx+c)+dy = a\sin(bx + c) + d, the period TT in degrees is calculated as T=360∘∣b∣T = \frac{360^\circ}{|b|}. Given y=5sin⁡(23x)−2y = 5\sin\left(\frac{2}{3}x\right) - 2, the value of bb is 23\frac{2}{3}. Dividing 360∘360^\circ by 23\frac{2}{3} gives 540∘540^\circ.

Step-by-Step Solution

1
Identify the coefficient bb of xx from the standard sine form y=asin⁡(bx+c)+dy = a\sin(bx + c) + d.
b=23b = \frac{2}{3}
The coefficient of xx determines the angular frequency and affects the horizontal compression or stretch of the graph.
2
State the period formula in degrees for a sine function.
T=360∘∣b∣T = \frac{360^\circ}{|b|}
The standard sine function completes one full wavelength over 360∘360^\circ, so scaling the input by bb changes the period to 360∘b\frac{360^\circ}{b}.
3
Substitute b=23b = \frac{2}{3} into the formula and evaluate.
T=360∘23=360∘×32=540∘T = \frac{360^\circ}{\frac{2}{3}} = 360^\circ \times \frac{3}{2} = 540^\circ
Dividing by a fraction is performed by multiplying by its reciprocal.

Key Concept

Period of Trigonometric Functions
Estimated Time:1m 30s
Question 9Question →

Which set contains all the solutions to the trigonometric equation sin⁡2x=cos⁡x\sin 2x = \cos x for 0∘≤x≤180∘0^\circ \le x \le 180^\circ?

Show answer & explanation

Answer: {30∘,90∘,150∘}\{30^\circ, 90^\circ, 150^\circ\}

Answer

The correct set of solutions is \{30^\circ, 90^\circ, 150^\circ\}.
Using the identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x, the equation becomes 2sin⁡xcos⁡x−cos⁡x=02\sin x \cos x - \cos x = 0. Factoring out cos⁡x\cos x gives cos⁡x(2sin⁡x−1)=0\cos x(2\sin x - 1) = 0. Setting each factor to zero yields cos⁡x=0\cos x = 0 (giving x=90∘x = 90^\circ) and sin⁡x=12\sin x = \frac{1}{2} (giving x=30∘x = 30^\circ and x=150∘x = 150^\circ). Thus, the complete set of solutions in the given interval is \{30^\circ, 90^\circ, 150^\circ\}.

Step-by-Step Solution

1
Apply the double-angle identity for sine.
Substitute sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x into the equation to get 2sin⁡xcos⁡x=cos⁡x2\sin x \cos x = \cos x.
This expresses the equation in terms of single angle xx.
2
Rearrange and factor the equation.
2sin⁡xcos⁡x−cos⁡x=0  ⟹  cos⁡x(2sin⁡x−1)=02\sin x \cos x - \cos x = 0 \implies \cos x(2\sin x - 1) = 0.
Factoring prevents losing valid roots that occur when a variable factor equals zero.
3
Set each factor to zero and solve for xx in the interval 0∘≤x≤180∘0^\circ \le x \le 180^\circ.
First factor: cos⁡x=0  ⟹  x=90∘\cos x = 0 \implies x = 90^\circ.
Second factor: 2sin⁡x−1=0  ⟹  sin⁡x=12  ⟹  x=30∘2\sin x - 1 = 0 \implies \sin x = \frac{1}{2} \implies x = 30^\circ or x=180∘−30∘=150∘x = 180^\circ - 30^\circ = 150^\circ.
Finding all principal and secondary angles within the specified domain.
4
Combine all valid solutions into a set.
x∈{30∘,90∘,150∘}x \in \{30^\circ, 90^\circ, 150^\circ\}.
All three values satisfy the original equation and lie within 0∘≤x≤180∘0^\circ \le x \le 180^\circ.

Key Concept

Solving trigonometric equations using identities and factoring
Estimated Time:1m 30s
Question 10Question →

Which of the following sets contains all values of xx in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ that satisfy the trigonometric equation 3sin⁡x+cos⁡x=0\sqrt{3}\sin x + \cos x = 0?

Show answer & explanation

Answer: 150∘ and 330∘150^\circ \text{ and } 330^\circ

Answer

150∘ and 330∘150^\circ \text{ and } 330^\circ
The given equation 3sin⁡x+cos⁡x=0\sqrt{3}\sin x + \cos x = 0 simplifies to tan⁡x=−13\tan x = -\frac{1}{\sqrt{3}}. Since tangent is negative in the second and fourth quadrants with a reference angle of 30∘30^\circ, the solutions in the domain 0∘≤x≤360∘0^\circ \le x \le 360^\circ are 180∘−30∘=150∘180^\circ - 30^\circ = 150^\circ and 360∘−30∘=330∘360^\circ - 30^\circ = 330^\circ.

Step-by-Step Solution

1
Rearrange the trigonometric equation into single ratio form
3sin⁡x=−cos⁡x  ⟹  sin⁡xcos⁡x=−13  ⟹  tan⁡x=−13\sqrt{3}\sin x = -\cos x \implies \frac{\sin x}{\cos x} = -\frac{1}{\sqrt{3}} \implies \tan x = -\frac{1}{\sqrt{3}}
Dividing both sides by cos⁡x\cos x converts the sum of sine and cosine terms into a simple tangent equation.
2
Determine the reference angle
Reference angle α=30∘\text{Reference angle } \alpha = 30^\circ
The acute angle whose tangent is 13\frac{1}{\sqrt{3}} is 30∘30^\circ.
3
Identify the quadrants and find all solutions in 0∘≤x≤360∘0^\circ \le x \le 360^\circ
x=180∘−30∘=150∘x = 180^\circ - 30^\circ = 150^\circ (Quadrant II) and x=360∘−30∘=330∘x = 360^\circ - 30^\circ = 330^\circ (Quadrant IV)
The tangent function is negative in Quadrants II and IV.

Key Concept

Solving simple trigonometric equations by reducing to a basic ratio and finding all solutions within a given domain.
Estimated Time:1m 30s
Question 11Question →

Find the value of xx, in degrees, for 0∘≤x≤90∘0^\circ \le x \le 90^\circ that satisfies the trigonometric equation sin⁡2x=cos⁡(x+30∘)\sin 2x = \cos(x + 30^\circ).

Show answer & explanation

Answer: 20

Answer

The value of xx in the interval 0∘≤x≤90∘0^\circ \le x \le 90^\circ satisfying the equation is 20∘20^\circ.
Using the co-function identity cos⁡α=sin⁡(90∘−α)\cos \alpha = \sin(90^\circ - \alpha), we convert the right-hand side to sin⁡(90∘−(x+30∘))=sin⁡(60∘−x)\sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x). Equating the arguments gives 2x=60∘−x2x = 60^\circ - x, which simplifies to 3x=60∘3x = 60^\circ, yielding x=20∘x = 20^\circ.

Step-by-Step Solution

1
Apply the co-function trigonometric identity
cos⁡(x+30∘)=sin⁡(90∘−(x+30∘))=sin⁡(60∘−x)\cos(x + 30^\circ) = \sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x)
Converting cosine to sine allows direct comparison of sine functions on both sides of the equation.
2
Set up the equation equating the angle expressions
2x=60∘−x2x = 60^\circ - x
Since sin⁡(2x)=sin⁡(60∘−x)\sin(2x) = \sin(60^\circ - x) and x∈[0∘,90∘]x \in [0^\circ, 90^\circ], equating the principal angle arguments gives the primary solution.
3
Solve the linear equation for xx
3x=60∘  ⟹  x=20∘3x = 60^\circ \implies x = 20^\circ
Adding xx to both sides gives 3x=60∘3x = 60^\circ, and dividing by 3 yields x=20∘x = 20^\circ.

Key Concept

Co-function identities and simple trigonometric equations
Question 12Question →

For the domain 0∘≤x≤180∘0^\circ \le x \le 180^\circ, what is the complete solution set to the trigonometric equation 4sin⁡2x−3=04\sin^2 x - 3 = 0?

Show answer & explanation

Answer: {60∘,120∘}\{60^\circ, 120^\circ\}

Answer

{60∘,120∘}\{60^\circ, 120^\circ\}
Solving the equation 4sin⁡2x−3=04\sin^2 x - 3 = 0 gives sin⁡2x=34\sin^2 x = \frac{3}{4}, which simplifies to sin⁡x=32\sin x = \frac{\sqrt{3}}{2} within the interval 0∘≤x≤180∘0^\circ \le x \le 180^\circ. The principal angle is 60∘60^\circ. Since sine is positive in both the first and second quadrants, the second valid angle in the domain is 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ, giving the solution set {60∘,120∘}\{60^\circ, 120^\circ\}.

Step-by-Step Solution

1
Isolate the trigonometric term
sin⁡2x=34\sin^2 x = \frac{3}{4}
Rearrange 4sin⁡2x−3=04\sin^2 x - 3 = 0 by adding 3 to both sides and dividing by 4.
2
Take the square root of both sides
sin⁡x=±32\sin x = \pm \frac{\sqrt{3}}{2}
Taking the square root yields both positive and negative ratios.
3
Apply the domain restriction 0∘≤x≤180∘0^\circ \le x \le 180^\circ
sin⁡x=32\sin x = \frac{\sqrt{3}}{2}
The sine function is non-negative in the first and second quadrants (0∘≤x≤180∘0^\circ \le x \le 180^\circ), so the negative root has no solutions in this interval.
4
Determine the angles for xx
x = 60^\circ \text{ and } x = 180^\circ - 60^\circ = 120^\circ
The reference angle is 60∘60^\circ because sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}. In Quadrant II, the corresponding angle is 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ.

Key Concept

Solving Quadratic Trigonometric Equations
Question 13Question →

Find the sum of all values of xx (in degrees) in the interval 0∘≤x≤180∘0^\circ \le x \le 180^\circ that satisfy the trigonometric equation cos⁡(3x−45∘)=−22\cos(3x - 45^\circ) = -\frac{\sqrt{2}}{2}.

Show answer & explanation

Answer: 330

Answer

The sum of all values of xx satisfying the equation in the domain 0∘≤x≤180∘0^\circ \le x \le 180^\circ is 330.
Transforming the domain 0∘≤x≤180∘0^\circ \le x \le 180^\circ gives −45∘≤3x−45∘≤495∘-45^\circ \le 3x - 45^\circ \le 495^\circ. The angles within this range where the cosine value equals −22-\frac{\sqrt{2}}{2} are 135∘135^\circ, 225∘225^\circ, and 495∘495^\circ. Solving 3x−45∘3x - 45^\circ for each of these angles gives x=60∘x = 60^\circ, 90∘90^\circ, and 180∘180^\circ. Summing these three roots yields 330∘330^\circ.

Step-by-Step Solution

1
Determine the interval of the transformed angle θ=3x−45∘\theta = 3x - 45^\circ.
−45∘≤3x−45∘≤495∘-45^\circ \le 3x - 45^\circ \le 495^\circ
Applying the linear transformation 3x−45∘3x - 45^\circ to the given domain 0∘≤x≤180∘0^\circ \le x \le 180^\circ establishes the exact boundaries for the argument of the cosine function.
2
Find all values of θ\theta within [−45∘,495∘][-45^\circ, 495^\circ] satisfying cos⁡θ=−22\cos \theta = -\frac{\sqrt{2}}{2}.
θ∈{135∘,225∘,495∘}\theta \in \{135^\circ, 225^\circ, 495^\circ\}
Cosine is negative in Quadrants II and III. The reference angle is 45∘45^\circ, giving base solutions 135∘135^\circ and 225∘225^\circ. Adding 360∘360^\circ to 135∘135^\circ gives 495∘495^\circ, which lies exactly on the upper boundary.
3
Solve for xx by setting 3x−45∘3x - 45^\circ equal to each valid θ\theta.
x∈{60∘,90∘,180∘}x \in \{60^\circ, 90^\circ, 180^\circ\}
Isolating xx yields x=θ+45∘3x = \frac{\theta + 45^\circ}{3}. All three resulting values lie within [0∘,180∘][0^\circ, 180^\circ].
4
Sum the valid solution values.
60^\circ + 90^\circ + 180^\circ = 330^\circ
The problem asks specifically for the sum of all solution angles in degrees.

Key Concept

Solving multi-angle trigonometric equations with phase shifts across a specified domain
Question 14Question →

What is the period, in degrees, of the trigonometric function y=7sin⁡(5x)−2y = 7\sin(5x) - 2?

Show answer & explanation

Answer: 72

Answer

The period of the trigonometric function is 72 degrees.
For any function of the form y=Asin⁡(Bx)+Dy = A\sin(Bx) + D, the period TT in degrees is calculated using T=360∘∣B∣T = \frac{360^\circ}{|B|}. For the given equation y=7sin⁡(5x)−2y = 7\sin(5x) - 2, the value of BB is 5. Substituting this into the formula gives T=360∘5=72∘T = \frac{360^\circ}{5} = 72^\circ.

Step-by-Step Solution

1
Identify the coefficient BB of the variable xx in the given function y=7sin⁡(5x)−2y = 7\sin(5x) - 2.
Here, A=7A = 7, B=5B = 5, and D=−2D = -2.
The period of a sine function depends on the angular frequency parameter BB multiplying the input variable xx.
2
Apply the standard formula for finding the period TT of a sine function in degrees: T=360∘∣B∣T = \frac{360^\circ}{|B|}.
T=360∘5=72∘T = \frac{360^\circ}{5} = 72^\circ.
Dividing the standard full revolution of 360∘360^\circ by the multiplier 55 determines the angle needed for one full cycle.

Key Concept

Period of a Trigonometric Graph
Estimated Time:45s
Question 15Question →

Find the smallest positive value of θ\theta, in degrees, that satisfies the trigonometric equation 2sin⁡(3θ−30∘)=32\sin(3\theta - 30^\circ) = \sqrt{3}.

Show answer & explanation

Answer: 30

Answer

The smallest positive angle θ\theta is 30∘30^\circ.
To find the smallest positive value of θ\theta, first isolate the sine function by dividing both sides by 2 to obtain sin⁡(3θ−30∘)=32\sin(3\theta - 30^\circ) = \frac{\sqrt{3}}{2}. The smallest positive angle with a sine of 32\frac{\sqrt{3}}{2} is 60∘60^\circ. Setting 3θ−30∘=60∘3\theta - 30^\circ = 60^\circ yields 3θ=90∘3\theta = 90^\circ, which gives θ=30∘\theta = 30^\circ.

Step-by-Step Solution

1
Isolate the trigonometric ratio
sin⁡(3θ−30∘)=32\sin(3\theta - 30^\circ) = \frac{\sqrt{3}}{2}
Dividing both sides of 2sin⁡(3θ−30∘)=32\sin(3\theta - 30^\circ) = \sqrt{3} by 2 simplifies the equation into standard form.
2
Determine the primary angle solution
3θ−30∘=60∘3\theta - 30^\circ = 60^\circ
The smallest positive angle whose sine equals 32\frac{\sqrt{3}}{2} is 60∘60^\circ.
3
Solve the linear equation for θ\theta
θ=30∘\theta = 30^\circ
Adding 30∘30^\circ to both sides gives 3θ=90∘3\theta = 90^\circ, and dividing by 3 yields θ=30∘\theta = 30^\circ.

Key Concept

Solving Trigonometric Equations with Linear Argument Transformations
Question 16Question →

A trigonometric function is defined as f(x)=asin⁡(bx)+cf(x) = a \sin(b x) + c, where a>0a > 0 and b>0b > 0. The graph of y=f(x)y = f(x) has a maximum value of 88, a minimum value of −2-2, and a period of 120∘120^\circ. What is the value of a+b+ca + b + c?

Show answer & explanation

Answer: 11

Answer

The value of a+b+ca + b + c is 11.
Using the maximum (a+c=8a + c = 8) and minimum (−a+c=−2-a + c = -2), solving simultaneously gives a=5a = 5 and c=3c = 3. Using the period formula T=360∘b=120∘T = \frac{360^\circ}{b} = 120^\circ, we obtain b=3b = 3. Adding these values together yields 5+3+3=115 + 3 + 3 = 11.

Step-by-Step Solution

1
Set up a system of linear equations for the amplitude and vertical shift from the maximum and minimum bounds
a=5a = 5 and c=3c = 3
Since sin⁡(bx)\sin(bx) ranges from −1-1 to 11, the maximum is a(1)+c=8a(1) + c = 8 and minimum is a(−1)+c=−2a(-1) + c = -2. Solving these simultaneous equations gives c=3c = 3 and a=5a = 5.
2
Calculate the frequency coefficient bb using the period formula
b=3b = 3
For a sine curve specified in degrees, the period is T=360∘bT = \frac{360^\circ}{b}. Substituting T=120∘T = 120^\circ yields b=3b = 3.
3
Compute the requested sum a+b+ca + b + c
a+b+c=11a + b + c = 11
Summing the calculated parameters 5+3+3=115 + 3 + 3 = 11.

Key Concept

Determining parameters of a trigonometric graph from amplitude, vertical shift, and period.
Question 17Question →

Determine the number of distinct solutions to the trigonometric equation 2cos⁡2θ+sin⁡θ−1=02\cos^2 \theta + \sin \theta - 1 = 0 within the interval 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.

Show answer & explanation

Answer: 3

Answer

The total number of distinct solutions in the given interval is 3.
Substituting cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta yields the quadratic equation 2sin⁡2θ−sin⁡θ−1=02\sin^2 \theta - \sin \theta - 1 = 0. Factoring gives sin⁡θ=1\sin \theta = 1 and sin⁡θ=−12\sin \theta = -\frac{1}{2}. Within 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ, sin⁡θ=1\sin \theta = 1 gives one solution (90∘90^\circ), while sin⁡θ=−12\sin \theta = -\frac{1}{2} gives two solutions (210∘210^\circ and 330∘330^\circ). In total, there are 3 distinct solutions.

Step-by-Step Solution

1
Substitute the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta into the original equation
2(1−sin⁡2θ)+sin⁡θ−1=02(1 - \sin^2 \theta) + \sin \theta - 1 = 0
Converting the equation into a single trigonometric ratio allows for algebraic solving.
2
Simplify and arrange into quadratic form
2sin⁡2θ−sin⁡θ−1=02\sin^2 \theta - \sin \theta - 1 = 0
This puts the expression into standard quadratic form au2+bu+c=0au^2 + bu + c = 0 where u=sin⁡θu = \sin \theta.
3
Factor the quadratic equation
(2sin⁡θ+1)(sin⁡θ−1)=0(2\sin \theta + 1)(\sin \theta - 1) = 0
Factoring determines the roots for sin⁡θ\sin \theta.
4
Solve for possible values of sin⁡θ\sin \theta
sin⁡θ=1\sin \theta = 1 or sin⁡θ=−12\sin \theta = -\frac{1}{2}
By the zero-product property, at least one factor must equal zero.
5
Find the angles for each ratio in the interval 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ
θ=90∘,210∘,330∘\theta = 90^\circ, 210^\circ, 330^\circ
sin⁡θ=1\sin \theta = 1 has one solution (90∘90^\circ) and sin⁡θ=−0.5\sin \theta = -0.5 has two solutions in the 3rd and 4th quadrants (210∘210^\circ and 330∘330^\circ).
6
Count the solutions
3
There are 3 distinct values of θ\theta satisfying the condition.

Key Concept

Solving quadratic trigonometric equations using basic identities and quadrant analysis
Estimated Time:1m 30s
Question 18Question →

Find the acute angle θ\theta, in degrees, that satisfies the trigonometric equation 3tan⁡θ−3=0\sqrt{3}\tan \theta - 3 = 0.

Show answer & explanation

Answer: 60

Answer

The acute angle θ\theta is 60∘60^\circ.
Rearranging the equation 3tan⁡θ−3=0\sqrt{3}\tan \theta - 3 = 0 gives 3tan⁡θ=3\sqrt{3}\tan \theta = 3, so tan⁡θ=33=3\tan \theta = \frac{3}{\sqrt{3}} = \sqrt{3}. For an acute angle (0∘<θ<90∘0^\circ < \theta < 90^\circ), the angle with a tangent equal to 3\sqrt{3} is 60∘60^\circ.

Step-by-Step Solution

1
Isolate the trigonometric ratio tan⁡θ\tan \theta
tan⁡θ=3\tan \theta = \sqrt{3}
Add 33 to both sides and divide by 3\sqrt{3}, giving 33=3\frac{3}{\sqrt{3}} = \sqrt{3}.
2
Determine the value of the acute angle θ\theta
θ=60∘\theta = 60^\circ
From special angle exact values, tan⁡(60∘)=3\tan(60^\circ) = \sqrt{3}.

Key Concept

Solving Simple Trigonometric Equations
Question 19Question →

What is the maximum value of the trigonometric function y=3sin⁡x+2y = 3\sin x + 2?

Show answer & explanation

Answer: 5

Answer

The maximum value of the function is 5.
The basic sine function sin⁡x\sin x reaches a maximum value of 11. Substituting sin⁡x=1\sin x = 1 into y=3sin⁡x+2y = 3\sin x + 2 yields y=3(1)+2=5y = 3(1) + 2 = 5.

Step-by-Step Solution

1
Identify the maximum value of the sine term
The range of sin⁡x\sin x is [−1,1][-1, 1], so its maximum value is 11.
The sine function oscillates between a minimum of −1-1 and a maximum of 11 for all real numbers xx.
2
Calculate the maximum value of the transformed function
ymax=3(1)+2=5y_{\text{max}} = 3(1) + 2 = 5.
Multiplying by the positive amplitude coefficient 33 scales the peak to 33, and adding the vertical shift of 22 raises the peak to 55.

Key Concept

Maximum and Minimum Values of Trigonometric Functions
Question 20Question →

At which values of xx within the domain 0∘≤x≤360∘0^\circ \le x \le 360^\circ do the graphs of f(x)=2sin⁡xf(x) = 2\sin x and g(x)=tan⁡xg(x) = \tan x intersect?

Show answer & explanation

Answer: 0∘,60∘,180∘,300∘,360∘0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

Answer

The graphs intersect at x=0∘,60∘,180∘,300∘,360∘x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ.
Equating 2sin⁡x=sin⁡xcos⁡x2\sin x = \frac{\sin x}{\cos x} yields sin⁡x(2cos⁡x−1)=0\sin x (2\cos x - 1) = 0. Setting sin⁡x=0\sin x = 0 gives x=0∘,180∘,360∘x = 0^\circ, 180^\circ, 360^\circ, and setting cos⁡x=12\cos x = \frac{1}{2} gives x=60∘,300∘x = 60^\circ, 300^\circ. Combining these gives the full solution set.

Step-by-Step Solution

1
Set the two trigonometric expressions equal to each other to find points of intersection.
2sin⁡x=tan⁡x2\sin x = \tan x
Intersection points occur where f(x)=g(x)f(x) = g(x).
2
Rewrite tan⁡x\tan x in terms of sine and cosine.
2sin⁡x=sin⁡xcos⁡x2\sin x = \frac{\sin x}{\cos x}, for x≠90∘,270∘x \neq 90^\circ, 270^\circ
Using the quotient identity tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} allows simplification.
3
Rearrange the equation and factor out sin⁡x\sin x.
2sin⁡xcos⁡x−sin⁡x=0  ⟹  sin⁡x(2cos⁡x−1)=02\sin x \cos x - \sin x = 0 \implies \sin x(2\cos x - 1) = 0
Factoring prevents losing solutions that occur when sin⁡x=0\sin x = 0.
4
Solve each factor separately within 0∘≤x≤360∘0^\circ \le x \le 360^\circ.
Factor 1: sin⁡x=0  ⟹  x=0∘,180∘,360∘\sin x = 0 \implies x = 0^\circ, 180^\circ, 360^\circ.
Factor 2: 2cos⁡x−1=0  ⟹  cos⁡x=12  ⟹  x=60∘,300∘2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = 60^\circ, 300^\circ.
Cosine is positive in the first and fourth quadrants.
5
Combine all valid solutions.
x=0∘,60∘,180∘,300∘,360∘x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ
All five values satisfy the original equation and lie within the given domain.

Key Concept

Solving trigonometric equations by factoring and finding all roots in a given domain
Estimated Time:2m 0s
Page 1 / 2Next
Trigonometric Graphs and Simple Equations Practice Questions — JAMB UTME | Examkin