Answer:
x = -π/2, π/6, 5π/6
Explanation:
You are given a quadratic equation in sin(x) and cos(x) and asked for the values of x that make it true. The domain of interest is given.
Trig identity
The trig identity ...
cos(x)² = 1 -sin(x)²
is useful for turning this equation into a quadratic in sin(x). Making use of it, the equation becomes ...
1 -sin(x)² -sin(x)² = sin(x)
Solution
Subtracting the left-side expression from both sides gives the standard-form equation ...
2sin(x)² +sin(x) -1 = 0
Factoring this equation, we have ...
(2sin(x) -1)(sin(x) +1) = 0
The values of x that make these factors zero are ...
2sin(x) -1 = 0 ⇒ sin(x) = 1/2 ⇒ x = π/6 or 5π/6
sin(x) +1 = 0 ⇒ sin(x) = -1 ⇒ x = -π/2
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Additional comment
Sometimes the quadratic nature of the equation is easier to see if you make a substitution for the trig function.
sin(x) = z ⇒ 2z² +z -1 = 0 ⇒ (2z -1)(z +1) = 0 ⇒ z = 1/2 or -1
At this point, you need to find the values of x that correspond to these values of z. (In short, you're not quite done yet when you find the solutions for z.)