Даю 50 баллов Помогите пожалуйста с уравнением

Question 1

Question 2

Решите задачу:

$\frac{2-cos4x+(4-\sqrt3)\, sin2x+2\, cos2x-2\sqrt3}{1+2sin(\frac{9\pi }{2}+2x)}=1\\\\\star \; sin(\frac{9\pi }{2}+2x)=sin(4\pi +\frac{\pi }{2}+2x)=sin(\frac{\pi }{2}+2x)=cos2x\; \; \star \\\\\\ODZ:\; 1+2cos2x\ne 0\; ,\; cos2x\ne -\frac{1}{2}\; ,\; 2x\ne \pm arccos(-\frac{1}{2})+2\pi n\\\\2x\ne \pm (\pi -\frac{\pi }{3})+2\pi n\; ,\; 2x\ne \pm \frac{2\pi }{3}+2\pi n\; ,\; \underline {x\ne \pm \frac{\pi }{3}+\pi n\; ,\; n\in Z}\\\\\\\frac{2-cos4x+(4-\sqrt3)\, sin2x+2\, cos2x-2\sqrt3x}{1+2cos2x}-1=0$

$\frac{2-cos4x+(4-\sqrt3)sin2x+2cos2x-2\sqrt3-(1+2cos2x)}{1+2cos2x}=0\\\\1-cos4x+(4-\sqrt3)sin2x-2\sqrt3=0\; \; ,\; \; [\; cos4x=1-2sin^22x\; ]\\\\2sin^22x+(4-\sqrt3)sin2x-2\sqrt3=0\\\\D=(4-\sqrt3)^2+16\sqrt3=(16-8\sqrt3+3)+16\sqrt3=16+8\sqrt3+3=(4+\sqrt3)^2\\\\sin2x=\frac{-(4-\sqrt3)\pm \sqrt{(4+\sqrt3)^2}}{2\cdot 2}=\frac{-4+\sqrt3\pm (4+\sqrt3)}{4}=\left [{{\frac{\sqrt3}{2}} \atop {-2}} \right. \\\\a)\; \; sin2x=-2<-1\; \; \to \; \; x\in \varnothing \; ,t.k.\; |sinx|\leq 1\; ;$

$b)\; \; sin2x=\frac{\sqrt3}{2}\; ,\; \; 2x=\left [ {{\frac{\pi }{3}+2\pi m\; ,\; m\in Z } \atop {\frac{2\pi }{3}+2\pi m,\; m\in z}} \right. \; \; ,\; \; x=\left [ {{\frac{\pi}{6}+\pi m,\; m\in Z} \atop {\frac{\pi}{3}+\pi m\; \notin ODZ}} \right.\\\\Otvet:\; \; x=\frac{\pi }{6}+\pi m\; ,\; m\in Z\; .$