Question

sin2x+2sin*в_квадрате*x=0

Answer 1

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