Question

Решите пожалуйста тригонометрическое уравнение:

12sin^2x + 20cosx - 19 = 0

Answer 1

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