Question

cos(p/2-x)-sin3x+sin5x=0

Answer 1

$cos( dfrac{ pi }{2}-x)-sin3x+sin5x=0 \ sinx-sin3x+sin5x=0 \ 2sin3xcos2x-sin3x=0 \ sin3x(2cos2x-1)=0 \ \ sin3x=0 \ 3x= pi k \ x= dfrac{ pi k}{3}; k in Z \ \ 2cos2x-1=0 \ cos2x= dfrac{1}{2} \ 2x=б dfrac{ pi }{3}+2 pi k \ x=б dfrac{ pi }{6}+ pi k ; k in Z$

Ответ: $left[begin{array}{I} x= dfrac{ pi k}{3}; k in Z \ x=б dfrac{ pi }{6}+ pi k ; k in Z end{array}}$