Решите уравнение 6sin^2 x+5cosx-2=0

[tex]6sin ^{2} x+5cosx-2=0\\6(1-cos ^{2} x)+5cosx-2=0\\6-6cos ^{2} x+5cosx-2=0\\4-6cos ^{2} x+5cosx=0\\-6cos ^{2} x+5cosx+4=0/*(-1)\\6cos ^{2} x-5cosx-4=0 \\cosx=t\\-1 \leq t \leq 1 \\ 6t ^{2} -5t-4=0\\D=25+4*4*6=25+96=121\\ \sqrt{D} =11\\t _{1} = \frac{5+11}{12} = \frac{16}{12} =1 \frac{4}{12} =1 \frac{1}{3} \\ \\ t _{2} = \frac{5-11}{12} = \frac{-6}{12} =- \frac{1}{2} \\\\ [/tex]

[tex]cosx=- \frac{1}{2} \\ x=+-arccos(- \frac{1}{2} )+2 \pi n\\x=+-( \pi -arccos \frac{1}{2} )+2 \pi n\\x=+-( \pi - \frac{ \pi }{3} )+2 \pi n\\x=+-( \frac{3 \pi - \pi }{3} )+2 \pi n\\x=+- \frac{2 \pi }{3} +2 \pi n[/tex]
n ∈Z