Решите уравнениеsin 2x ctg x=кореньиз3 cos x.

Решите задачу:

$sin2x*ctgx= \sqrt{3} cosx \\ \\ 2sinxcosx* \frac{cosx}{sinx} =\sqrt{3} cosx \\ \\\frac{2sinxcosx*cosx}{sinx} -\sqrt{3} cosx=0 \\ \\ 2cos^2x- \sqrt{3} cosx=0 \\ \\ cosx(2cosx- \sqrt{3} )=0 \\ \\ \left \{ {{cosx=0} \atop {2cosx- \sqrt{3} =0}} \right. \\ \\ \left \{ {{cosx=0} \atop {cosx= \frac{ \sqrt{3}}{2} }} \right. \\ \\ \left \{ {{x= \frac{ \pi } {2} + \pi n} \atop {x=б \frac{ \pi }{6}+2 \pi n }} \right. \\ \\ OTBET: \frac{ \pi } {2} + \pi n; б \frac{ \pi }{6}+2 \pi n; n \in Z$