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