Решите кто нибудь №6 б) буду признательна и благодарна

Решите задачу:

$\sqrt3sinx+cosx=\sqrt2\; |:2\\\\\frac{\sqrt3}{2}sinx+\frac{1}{2}cosx=\frac{\sqrt2}{2}\\\\cos\frac{\pi}{6}sinx+sin\frac{\pi}{6}cosx=\frac{\sqrt2}{2}\\\\sin(x+\frac{\pi}{6})=\frac{\sqrt2}{2}\\\\x+\frac{\pi}{6}=(-1)^{n}\frac{\pi}{4}+\pi n,n\in Z\\\\x=(-1)^{n}\frac{\pi}{4}-\frac{\pi}{6}+\pi n\\\\2.\; 1-cos\frac{x}{2}=tg\frac{x}{4}\\\\2sin^2\frac{x}{4}-tg\frac{x}{4}=0$

$sin\frac{x}{4}(2sin\frac{x}{4}-\frac{1}{cos\frac{x}{4}})=0\\\\a)\; sin\frac{x}{4}=0,\frac{x}{4}=\pi n,x=4\pi n,n\in Z\\\\b)\; \{2sin\frac{x}{4}cos\frac{x}{4}-1=0,cos\frac{x}{4}\ne 0\}\\\\sin\frac{x}{2}=1,\frac{x}{2}=\frac{\pi}{2}+\pi k,x=\pi +2\pi k,k\in Z\\\\\frac{x}{4}\ne \frac{\pi}{2}+\pi m,x\ne 2\pi +4\pi m,m\in Z\\\\Otvet:\; x=4\pi n,x=\pi +2\pi k;\; n,k\in Z$