Решите уравнение: 2cos квадрате x-cosx-1=0

[latex]2cos ^{2} -cosx-1=0[/latex] [latex]cosx=t[/latex] [latex]2t ^{2} -t-1=0[/latex] D = (-1)² - 4· 2·(-1) = 1 + 8 = 9 [latex] \sqrt{D} =3[/latex] [latex] t_{1} = \frac{1+3}{4} = \frac{4}{4}=1 [/latex] [latex]t _{2} = \frac{1-3}{4} = \frac{-2}{4}=- \frac{1}{2} [/latex] [latex] \left \{ {{cosx=1} \atop {cosx=- \frac{1}{2} }} \right. [/latex] [latex]x=2 \pi k,[/latex] k∈Z [latex]x= \frac{2 \pi }{3}+2 \pi n, [/latex] n∈Z