Решите уравнение: 2cos^2x + 4 = 3√3cos ((3п/2)+x)

Решите задачу:

$2Cos^{2}x+4=3\sqrt{3}Cos(\frac{3\pi }{2} +x)\\\\2Cos^{2}x+4-3\sqrt{3}Sinx=0\\\\2(1-Sin^{2}x)+4-3\sqrt{3}Sinx=0\\\\2-2Sin^{2}x+4-3\sqrt{3}Sinx=0\\\\2Sin^{2}x+3\sqrt{3} Sinx-6=0\\\\D=(3\sqrt{3})^{2}-4*2*(-6)=27+48=75=(5\sqrt{3})^{2}\\\\1)Sinx=\frac{-3\sqrt{3}+5\sqrt{3}}{4} =\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\\\\x=(-1)^{n}arc Sin\frac{\sqrt{3}}{2} +\pi n,n\in Z\\\\x=(-1)^{n}\frac{\pi }{3} +\pi n,n\in Z$

$2)Sinx=\frac{-3\sqrt{3}-5\sqrt{3}}{4}=-\frac{8\sqrt{3}}{4}=-2\sqrt{3}$