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Difficulty: HardTrigonometric Graphs and Simple Equations

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?

  1. 30∘,90∘,150∘30^\circ, 90^\circ, 150^\circAnswer
  2. B
    60∘,90∘,120∘60^\circ, 90^\circ, 120^\circ
  3. C
    30∘,150∘30^\circ, 150^\circ
  4. D
    60∘,300∘60^\circ, 300^\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
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Topics

Trigonometric Graphs and Simple Equations

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