Question

Корень 3 cos2x=sinxcosx

Answer 1

$sqrt{3}cos2x=sinxcosx$

$sqrt{3}cos2x=frac{1}{2}sin2x$

$2sqrt{3}cos2x-sin2x=0$

$cos2x(2sqrt{3}-tg2x)=0$

$x=frac{pi}{4}+frac{pi}{2}k,k in Z$

$x=frac{arctg(2sqrt{3})}{2}+frac{pi}{2}t,t in Z$