Помогите cos(3x-п/6)+cos(x+п/4)=0

Решите задачу:

$\displaystyle cos(3x- \frac{ \pi }{6})+cos(x+ \frac{ \pi }{4})=0\\2cos( \frac{3x- \pi /6+x+ \pi /4}{2})*cos( \frac{3x- \pi /6-x- \pi /4}{2})=0\\2cos(x+ \frac{ \pi }{24}) *cos(x- \frac{5x}{24})=0$

$\displaystyle cos(x+ \frac{ \pi }{24})=0\\x+ \frac{ \pi }{24}= \frac{ \pi }{2}+ \pi n; n\in Z\\x=- \frac{ \pi }{24}+ \frac{ \pi }{2}+ \pi n; n\in Z\\x= \frac{11 \pi }{24}+ \pi n; n\in Z$

$\displaystyle cos(x- \frac{5 \pi }{24})=0\\x- \frac{5 \pi }{24}= \frac{ \pi }{2}+ \pi n; n\in Z\\x= \frac{5 \pi }{24}+ \frac{ \pi }{2}+ \pi n; n\in Z\\x= \frac{17 \pi }{24}+ \pi n; n\in Z$