cos2x - cosx = 0cos^2 x - sin^2 x - cosx = 0cos^2 x - (1 - sin^2 x) - cosx = 02cos^2 x-1 - cosx = 0cosx*(2cos x - 1)= 1cos x = 1x= 2*пи*n
![\cos2x-\cos x=0\\ \cos2x=2\cos^2x-1\\ 2\cos^2x-1-\cos x=0\\ 2\cos^2x-\cos x-1=0\\ \cos x=t,\;\;\cos^2x=t^2,\;\;t\in[-1;1]\\ 2t^2-t-1=0\\ D=1+4\cdot8=9\\ t_1=1\\ t_2=-\frac12\\ \cos x=1\Rightarrow x=2\pi n,\;n\in\mathbb{Z}\\ \cos x=-\frac12\Rightarrow x=\frac{2\pi}3+2\pi n,\;n\in\mathbb{Z}](/tpl/images/0171/5855/d828b.png)
