Question

cosx*sin5x=cos2x*sin4x

Answer 1

$cosxsin5x=cos2xsin4x \ dfrac{1}{2}(sin4x+sin6x)= dfrac{1}{2}(sin2x+sin6x) \ sin4x-sin2x=0 \ sinxcos3x =0 \ \ sinx=0 \ x= pi k; k in Z \ \ cos3x=0 \ 3x= dfrac{ pi }{2}+ pi k \ x= dfrac{ pi }{6}+ dfrac{ pi k}{3}; k in Z$

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