cos4x+cos2x=0 решите уравнение

[latex]\cos4x+\cos2x=0\\ \cos^22x-\sin^22x+\cos2x=0\\ \cos^22x-(1-\cos^22x)+\cos2x=0\\ 2\cos^22x+\cos2x-1=0\\ \cos2x=t,\,\,\, |t|\le1\\ 2t^2+t-1=0\\ t_1=\frac{-1+3}4=\frac12\\ t_2=\frac{-1-3}4=-1\\ \\ \cos2x=\frac12\\ 2x=\pm\frac\pi3+2\pi k\\ x_1=\pm\frac\pi{6}+\pi k \\ \\ \cos2x=-1\\ 2x=\pi+2\pi n\\ x_2=\frac\pi2+\pi n [/latex]