-1+2cos^2+cos4x+√2 *cosx=0

Решите задачу:

$\underbrace {-1+2cos^2x}_{cos2x}+cos4x+\sqrt2cosx=0\\\\\star \; \; cos2x=cos^2x-sin^2x=cos^2x-(1-cos^2x)=2cos^2x-1\; \star \\\\cos2x+cos4x+\sqrt2cosx=0\\\\\star \; \; cos2x+cos4x=2cos \frac{2x+4x}{2}\cdot cos\frac{4x-2x}{2}=2cos3x\cdot cosx\; \star \\\\2cos3x\cdot cosx+\sqrt2\cdot cosx=0\\\\cosx(2cos3x+\sqrt2)=0\\\\1)\; cosx=0\; ,\; \; \underline {x=\frac{\pi}{2}+\pi n,\; n\in Z}\\\\2)\; \; cos3x=-\frac{\sqrt2}{2}\; ,\; \; 3x=\pm arccos(-\frac{\sqrt2}{2})+2\pi k,\; k\in Z \\\\3x=\pm (\pi -\frac{\pi}{4})+2\pi k=\pm \frac{3\pi}{4}+2\pi k,\; k\in Z$

$\underline {x=\pm \frac{\pi}{4}+ \frac{2\pi k}{3}\; ,\; k\in Z}$