ПОМОГИТЕ РЕШИТЬ ТРИГОНОМЕТРИЧЕСКОЕ УРАВНЕНИЕ ПОЖАЛУЙСТА!!!!

Решите задачу:

$\sqrt{10}cosx-\sqrt{4cosx-cos2x}=0\\\sqrt{10}cosx=\sqrt{4cosx-cos2x}\\cosx \geq0\\ (\sqrt{10}cosx) ^{2}=(\sqrt{4cosx-cos2x}) ^{2}\\10cos^{2}x=4cosx-cos2x\\10cos^{2}x=4cosx-(2cos^{2}x-1 )\\10cos^{2}x=4cosx-2cos^{2}x+1\\10cos^{2}x+2cos^{2}x-4cosx-1=0\\12cos^{2}x-4cosx-1=0\\cosx=a\\12a^{2}-4a-1=0\\D=16+48=64\\a_{1}=\frac{4+8}{24}=\frac{12}{24}=\frac{1}{2}\\a_{2}=\frac{4-8}{24}=-\frac{4}{24}=-\frac{1}{6}\\cosx=\frac{1}{2}\\x= \pm arccos(\frac{1}{2} ) + 2\pi n = \pm \frac{\pi}{3} + 2\pi n\\$