cos2x-2sin^2x

Question

Pre calculus problem

Answer 1

cos(2x)=1-2sin²(x),

cos(2x)-2sin²(x)=1-2sin²(x)-2sin²(x)=1-4sin²(x).

If cos(2x)-2sin²(x)=0, 1-4sin²(x)=0, (1-2sin(x))(1+2sin(x))=0,

so sin(x)=½ or -½. Therefore x=π/6 (30°), 5π/6 (150°), 7π/6 (210°), 11π/6 (330°).

There are other solutions found by adding multiples of 2π or 360° to each of these.