Cos^2 x + cos^2 2x= cos^2 3x

$cos^2(x)+cos^2(2x)=cos^2(3x)\\\\ cos^2(2x)=cos^2(3x)-cos^2(x)\\\\ cos^2(2x)=[cos(3x)-cos(x)]*[cos(3x)+cos(x)]\\\\ cos^2(2x)=[2*cos(2x)*cos(x)]*[-2*sin(2x)*sin(x)]\\\\ cos^2(2x)=-cos(2x)*2*sin(2x)*2*sin(x)*cos(x)\\\\ cos^2(2x)=-cos(2x)*2*sin(2x)*sin(2x)\\\\ cos^2(2x)+cos(2x)*2*sin^2(2x)=0\\\\ cos(2x)*[cos(2x)+2sin^2(2x)]=0\\\\ cos(2x)*[cos(2x)+2*(1-cos^2(2x))]=0\\\\ cos(2x)*[cos(2x)+2-2cos^2(2x)]=0\\\\ cos(2x)*[2cos^2(2x)-cos(2x)-2]=0\\\\ cos(2x)=t\ \ \ \ -1 \leq t \leq 1\\\\$

$\left \{ {{t*[2t^2-t-2]=0\\\\} \atop { -1\leq t \leq 1}} \right. \\\\ \left \{ {{t=0\ \ or\ \ 2t^2-t-2=0\ \ \ (D=17)\\\\} \atop { -1\leq t \leq 1}} \right. \\\\ \left \{ {{t=0\ \ or\ \ t=\frac{1\pm\sqrt{17}}{4}\\\\} \atop { -1\leq t \leq 1}} \right. \\\\ t=0\ \ or\ \ t=\frac{1-\sqrt{17}}{4}\\\\ cos(2x)=0\ \ or\ \ cos(2x)=\frac{1-\sqrt{17}}{4}\\\\ 2x=\frac{\pi}{2}+\pi n,\ n\in Z\ \ \ or\ \ \ 2x=\pm arccos(\frac{1-\sqrt{17}}{4})+2\pi k,\ k\in Z\\\\$

$x=\frac{\pi}{4}+\frac{\pi n}{2},\ n\in Z\ \ \ or\ \ \ x=\pm \frac{1}{2}arccos(\frac{1-\sqrt{17}}{4})+\pi k,\ k\in Z\\\\$

Ответ: $\frac{\pi}{4}+\frac{\pi n}{2},\ n\in Z\ \ ;\ \ x=\pm \frac{1}{2}arccos(\frac{1-\sqrt{17}}{4})+\pi k,\ k\in Z\\\\$