It looks like the equation is
-sin²(x) = cos(2x)
Recall the half-angle identity for sine:
sin²(x) = (1 - cos(2x))/2
Then the equation can be written as
-(1 - cos(2x))/2 = cos(2x)
Solve for cos(2x):
-1/2 + 1/2 cos(2x) = cos(2x)
-1/2 = 1/2 cos(2x)
cos(2x) = -1
On the unit circle, cos(y) = -1 when y = arccos(-1) = π. Since cosine has a period of 2π, more generally we have cos(y) = -1 for y = π + 2nπ where n is any integer. Then
2x = π + 2nπ
x = π/2 + nπ
In the interval [-π, π], you get two solutions x = -π/2 and x = π/2.
