Решите уравнение: 3sinx+1=cos2x

Cos (2x) + 3sinx+1 = 0

cos²x-sin²x+3sin (x) + cos²x+sin²x = 0

2cos²x+3sin (x) = 0

2 (1-sin²x) + 3sin (x) = 0

2-2sin²x+3sin (x) = 0

2sin²x-3sin (x) - 2 = 0

sin (x) = t

2t²-3t-2=0

d = 9+16 = 25

t1 = 2

t2=-1/2

sin (x) = 2 нет реш.

sin (x) = - 1/2

x=2π/3+2πk, k∈Z