Question

Difficulty: MediumTrigonometric Graphs and Simple Equations

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.

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
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