Решить уравнения:

Решите задачу:

$(1+ \sqrt{2} cos x)(1-4sinxcosx)=0 \\\ (1+ \sqrt{2} cos x)(1-2sin2x)=0 \\\ \left \{ {{1+ \sqrt{2} cos x=0} \atop {1-2sin2x=0}} \right. \\\ \left \{ {{cos x=- \frac{ \sqrt{2}}{2} } \atop {sin2x= \frac{1}{2} }} \right. \\\ \left \{ {{x=\pm \frac{ 3\pi}{4}+2\pi n } \atop {2x=(-1)^k \frac{\pi}{6}+\pi k }} \right. \\\ \left \{ {{x=\pm \frac{ 3\pi}{4}+2\pi n , n\in Z} \atop {x=(-1)^k \frac{\pi}{12}+ \frac{\pi k }{2}, k\in Z }} \right.$ $(1- \sqrt{2} cos x)(1+2sin2xcos2x)=0 \\\ (1- \sqrt{2} cos x)(1+sin4x)=0 \\\ \left \{ {{1- \sqrt{2} cos x=0} \atop {1+sin4x=0}} \right. \\\ \left \{ {{cos x= \frac{ \sqrt{2} }{2} } \atop {sin4x=-1}} \right. \\\ \left \{ {{x=\pm \frac{ \pi}{4}+2\pi n } \atop {4x=- \frac{\pi}{2}+2\pi n }} \right. \\\ \left \{ {{x=\pm \frac{ \pi}{4}+2\pi n , n\in Z } \atop {x=- \frac{\pi}{8}+ \frac{\pi n}{2}, n\in Z }} \right.$