Вопрос в картинках...

$\sqrt{3}sin \frac{x}{4} -cos \frac{x}{4} =0$

$2( \frac{ \sqrt{3} }{2} sin \frac{x}{4} - \frac{1}{2} cos \frac{x}{4})=0$

$\frac{ \sqrt{3} }{2} sin \frac{x}{4} - \frac{1}{2} cos \frac{x}{4}=0$

$\frac{ \sqrt{3} }{2} =cos \frac{ \pi }{6}, \frac{1}{2} =sin \frac{ \pi }{6}$

$cos \frac{ \pi }{6}sin \frac{x}{4} -sin \frac{ \pi }{6} cos \frac{x}{4} =0$

$sin( \frac{x}{4} - \frac{ \pi }{6} )=0$

$\frac{x}{4} - \frac{ \pi }{6} = \pi k,$ k∈Z

$\frac{x}{4} = \frac{ \pi }{6} +\pi k,$ k∈Z

$x= \frac{2 \pi }{3}+ \frac{ \pi }{4} k,$ k∈Z