Решите пожалуйста тригонометрическое уравнение: 12sin^2x + 20cosx - 19 = 0

Решите задачу:

$12sin ^{2} x+20cosx-19=0 \\ 12(1-cos ^{2} x)+20cosx-19=0 \\ 12-12cos ^{2} x+20cosx-19=0 \\ -12cos ^{2} x+20cosx-7=0 \\ cosx=t\ \ -1 \leq t \leq 1 \\ -12t ^{2} +20t-7=0 \\ 12t ^{2} -20t+7=0 \\ D=400-336=64 \\ \sqrt{D} =8 \\ t _{1} = \frac{20+8}{24} = \frac{28}{24} = \frac{14}{12} = \frac{7}{6} =1 \frac{1}{6} \\ t _{2} = \frac{20-8}{24} = \frac{12}{24} = \frac{1}{2} \\ cosx=0.5 \\ x=+- \frac{ \pi }{3} + \pi n,n\in Z$