Question

sin2x*cosx+sinx*cos2x=корень из 2/2

Answer 1

$displaystyle sin2xcosx+sinxcos2x=frac{sqrt2}{2}\sin(2x+x)=frac{sqrt2}{2}\sin(3x)=frac{sqrt2}{2}\3x_1=frac{pi}{4}+2pi n;3x_2=frac{3pi}{4}+2pi n;nin Z\x_1=frac{pi}{12}+frac{2pi n}{3};x_2=frac{3pi}{12}+frac{2pi n}{3};nin Z$