Решите 2,5, 8,9,10! очень надо

Решите задачу:

$2)\; \; cos6 \alpha +1=2cos^23 \alpha$

$Formyla:\\\\cos^2x=\frac{1+cos2x}{2}\; \; \to \; \; 1+cos2x=2cos^2x\; \; (x=3 \alpha )\\\\5)\; \; cos \alpha =\frac{2}{3}\; ,\; \; \frac{3\pi}{2}\ \textless \ \alpha \ \textless \ 2\pi \\\\sin \beta =\frac{1}{3}\; ,\; \; \frac{\pi}{2}\ \textless \ \beta \ \textless \ \pi \\\\\\sin \alpha =-\sqrt{1-cos^2 \alpha }=-\sqrt{1-\frac{4}{9}}=-\frac{\sqrt5}{3}\\\\cos \beta =-\sqrt{1-sin^2 \beta }=-\sqrt{1-\frac{1}{9}}=-\frac{2\sqrt2}{3}\\\\cos2 \beta =cos^2 \beta -sin^2 \beta =\frac{8}{9}-\frac{5}{9}=\frac{1}{3}$

$sin2 \beta =2sin \beta \cdot cos \beta =2\cdot \frac{1}{3} \cdot \frac{-2\sqrt2}{3} =-\frac{4\sqrt2}{9}$

$cos( \alpha -2 \beta )=cos \alpha \cdot cos2 \beta +sin \alpha \cdot sin2 \beta =$

$= \frac{2}{3} \cdot \frac{1}{3} +\frac{\sqrt5}{3} \cdot \frac{4\sqrt2}{9}=\frac{2}{9}+ \frac{4\sqrt{10}}{27}= \frac{6+4\sqrt{10}}{27}= \frac{2(3+2\sqrt{10})}{27}$