Use trigonometric identities to solve each equation within the given domain. –sin2(x) = cos(2x) from [–π, π]. PLEASE SHOW WORK!!!

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.

step-by-step explanation:

i am not sure sorry : c

50-60=10

step-by-step explanation: