Question

Sinx*cos2x=0
5sin^2x=2sinx

Answer 1

$1)sin*cos2x=0\ sinx=0 ili cos2x=0\ x=arcsin0 +pi n, neZ 2x=frac{pi}{2}+pi n, neZ\ x=pi n, neZ x=frac{pi}{4}+frac{pi n}{2}, neZ$

$2) 5sin^2x=2sinx\ frac{5sin^2x}{sinx}=2\ 5sinx=2\ sinx=frac25\ x=(-1)^n arcsinfrac25+pi n, neZ$