Помогите пожалуйста!

Решите задачу:

$\frac{1+\sqrt{3} }{2}*Sin2x=(\sqrt{3}-1)Cos^{2}x+1\\\\\frac{1+\sqrt{3}}{2}*2SinxCosx=(\sqrt{3}-1)Cos^{2}x+Sin^{2}x+Cos^{2}x\\\\(1+\sqrt{3})SinxCosx=(\sqrt{3}-1+1)Cos^{2}x+Sin^{2}x\\\\Sin^{2}x-(1+\sqrt{3})SinxCosx+\sqrt{3}Cos^{2}x=0|:Cos^{2}x,Cosx\neq0\\\\tg^{2}x-(1+\sqrt{3})tgx+\sqrt{3}=0\\\\D=(1+\sqrt{3})^{2}-4*\sqrt{3}=1+2\sqrt{3}+3-4\sqrt{3}=4-2\sqrt{3}=(1-\sqrt{3})^{2} \\\\tgx_{1}=\frac{1+\sqrt{3}+1-\sqrt{3}}{2}=1$

$x_{1} =arctg1+\pi n,n\in Z\\\\x_{1}=\frac{\pi }{4} +\pi n,n\in Z\\\\tgx_{2}=\frac{1+\sqrt{3}-1+\sqrt{3}}{2}=\frac{2\sqrt{3} }{2}=\sqrt{3}\\\\x_{2}=arctg\sqrt{3} +\pi n,n\in Z\\\\x_{2}=\frac{\pi }{3}+\pi n,n\in Z$