Verified answer

2sin^2x+cosx-1=0

(sin^2x +cos^2x = 1 => sin^2x = 1 - cos^2x)

2*(1 - cos^2x) +cosx-1=0

2cos^2x-cosx-1 = 0

t = cosx, t∈[-1;1]

2t^2 - t - 1 = 0

D = 1+8=9

t1= (1+3)/4=1

t2= (1-3)/4=-0.5

cosx=1                    cosx=-0.5

xn= 2Пn, n∈Z        xk=±arccos(-0.5) + 2Пk, k∈Z

                                 xk=±2П/3 + 2Пk, k∈Z

Ответ: xn= 2Пn, n∈Z; xk=±2П/3 + 2Пk, k∈Z