Question

–sin2(x) = cos(2x) from [–π,π]

I need help on this

Answer 1

If the equation is

- sin²(x) = cos(2x)

rewrite the left side using the Pythagorean identity,

sin²(x) = 1 - cos²(x)

and the right side using the double angle identity for cosine,

cos(2x) = 2 cos²(x) - 1

Then the equation transforms to

- (1 - cos²(x)) = 2 cos²(x) - 1

Solve for cos²(x) :

-1 + cos²(x) = 2 cos²(x) - 1

cos²(x) = 0

Take the square root of both sides to get

cos(x) = 0

Solve for x :

x = cos⁻¹(0) + 2nπ or x = cos⁻¹(0) - π + 2nπ

x = π/2 + 2nπ or x = -π/2 + 2nπ

where n is any integer. We get solutions in the interval [-π, π] when n = 0, giving the solutions

x = π/2 or x = -π/2