From the compound angle formula of cosine and the trigonometric identity of sin²x + cos²x = 1, cos2x = cos²x-sin²x = (1-sin²x)-sin²x = 1-2sin²x.** So, the given equation can be rewritten as follows: sinx = 1-sin²x ··· Eq.1
Let sinx = X, so the Eq.1 can be rewritten: X = 1-2X². Put this into the standard form of quadratic equation: 2X + X-1 = 0. To evaluate X, use the quadratic formula; x = (-b ± √(b²-4ac)) / 2a, where a = 2, b = 1 and c =-1. So, X = (-1 ± √9) / 4, we have: X = 1/2 or X = -1.
Evaluate x using the unit circle. When X = 1/2, sinx = 1/2. So, x = (2n + 1/6)π or, (2n + 5/6)π. And when X = -1, sinx = -1. So, x = (2n + 3/2)π (n: any integers)
Therefore, when x=(2n+1/6)π or (2n+5/6)π, sinx=cos2x=1/2. When x=(2n+3/2)π, sinx=cosx= -1
** ,or you can use the double-angle formula of cosine: cos2x = 1-2sin²x.