Question

Решите уравнение
√(1-cos2x)=sin2x

Answer 1

$( sqrt{1-cos2x} )^2=(sin2x)^2 \ 1-cos2x=sin^22x \ sin^22x+cos2x-1=0 \ 1-cos^22x+cos2x-1=0 \ cos^22x-cos2x=0 \ cos2x(cos2x-1)=0 \ \ cos2x=0 \ cos2x-1=0$

$cos2x=0 \ 2x= frac{ pi }{2}+ pi k, k in Z(:2) \ x= frac{ pi }{4}+ frac{ pi }{2}k, k in Z$

$cos2x=1 \ 2x=2 pi k, k in Z(:2) \ x= pi k, k in Z$