12cos^2 x - 20cos x + 7=0 решить уравнение

[latex]12cos ^{2} x-20cosx+7=0\\cosx=t\\12t ^{2} -20t+7=0\\D=400-4*7*12=400-336=64\\ \sqrt{D} =8\\t _{1} = \frac{20+8}{24} = \frac{28}{24} = \frac{7}{6} =1 \frac{1}{6} \\t _{2} = \frac{20-8}{24} = \frac{12}{24} = \frac{1}{2} \\ cosx= \frac{1}{2} \\x=+- \frac{ \pi }{3} +2 \pi n[/latex] n ∈Z