Question

sin2x*cosx+sinx*cos2x=корень из 2/2

Answer 1

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