Решить уравнение 2 sin^2 x-cos x-1=0

Sin²x + cos²x=1 ⇒sin²x=1-cos²x

2 sin ² x - cos x-1=0

2 (1-cos²x) - cos x-1=0

2-2cos²x - cos x-1=0

-2cos²x - cos x+1=0

2cos² (x) + cos (x) - 1=0

cos (x) = t

2t²+t-1=0

D=1+4=9 √D=3

t = (-1+3) / 4=1/2 cos (x) = 1/2 x = + / - π/3 + 2πn π∈Z

t = (-1-3) / 4 = - 1 cos (x) = - 1 x=π+2πn n∈Z