Pre Cal question

Question

Find all solutions; [0,2pi) with equation 2cos(x)-sin(2x)=0

Answer 1

0 = 2 cos(x) - sin(2x)

= 2 cos - 2 sin cos

= 2 cos ( 1 - sin)

so cos=0 or 1 - sin=0

For 1 -sin=0, sin=1, so the solution is pi/2

For cos=0, the solution is pi/2, (3/2)pi

{ pi/2, (3/2) pi}