Question

sin2x+2sin*в_квадрате*x=0

Answer 1

$sin2x+2sin^2x=0 \ 2sinxcosx+2sin^2x=0 \ 2sinx(cosx+sinx)=0 \ \ sinx=0 \ x= pi k , k in z \ \ cosx+sinx=0 \ cosx=-sinx (:cosx neq 0) \ tgx=-1 \ x=- frac{ pi }{4} + pi k , k in Z$