помогите с математикой плиз! 2sin^2 x + sin x-1=0 если с низу решите, получите лучший ответ

4 sin ^2 x + 11 sin x -3 = 0
6cos ^2 x + cosx-1=0
8sin^2 x - cos x +1 = 0

Question

4 sin ^2 x + 11 sin x -3 = 0 
6cos ^2 x + cosx-1=0 
8sin^2 x - cos x +1 = 0

Катюшка · Accepted Answer

2(sinx)^2 + sin x – 1 = 0, sinx = 1/2, x = (– 1)^k*(π/6) + πk, k∈Z
4(sinx)^2 + 11sin x – 3 = 0, sinx = 1/4, x = (– 1)^k*arcsin(1/4) + πk, k∈Z
6(cosx)^2 + cosx – 1 = 0, cosx = – 1/2, cosx = 1/3, x1 = ±2π/3 + 2πk, x2 = ±arccos(1/3)+ 2πk, k∈Z
8(sinx)^2 – cos x + 1 = 0, –8(cosx)^2 – cos x + 9 = 0, cosx = 1, x = 2πk, k∈Z

Irina Klimokhina · Answer

Ответ
2sin^2 x + sin x-1 = 0
sin^2 = t
2t^2 + t - 1 = 0
t1 = -2 => sin^2 x = -2 - исключается,
т. к. -1 =< sin x =< =1
t1 = 1 => sin x = 1 => x = п/2 + 2п*n
Остальные - аналогично.