Question

cos2x-2sinxcos2x=0

Answer 1

$cos2x-2sinxcos2x=0\cos2x(1-2sinx)=0\cos2x=0\2x= frac{ pi }{2} + pi n,nin Z \ x= frac{ pi }{4} + frac{ pi n}{2} ,nin Z \ \ 1-2sinx=0\-2sinx=-1 \ sinx= frac{1}{2} \ x=(-1) ^{n} frac{ pi }{6} + pi k,kin Z$

Answer 2

cos2x-2sinxcos2x=0
cos2x(1-2sinx)=0
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈Z
1-2sinx=0⇒2sinx=1⇒sinx=1/2⇒x=(-1)^n*π/6+πn,n∈Z