2cos(2x)-1=-3sin^2x; -2pi<x<2pi

Question

PLEASE ANSWER THE QUESTON!

Answer 1

2cos(2x)-1=-3sin²(x),

2(1-2sin²(x))-1+3sin²(x)=0,

2-4sin²(x)-1+3sin²(x)=0,

-sin²(x)+1=0,

sin(x)=±1, x=(2n+1)π/2, where n is an integer such that:

-2π<(2n+1)π/2<2π,

-4<2n+1<4, -5<2n<3, so -2≤n≤1, i.e., -2, -1, 0, 1.

x=-3π/2, -π/2, π/2, 3π/2.