-2sin²x + sin (pi/2 +x ) +1 = 0
-2sin²x + cosx +1 = 0
-2 ( 1 - cos²x ) + cosx +1 = 0
-2 + 2cos²x + cosx + 1 = 0
2cos²x + cosx - 1 =0
Пусть cosx = t, тогда
2t² + t - 1 = 0
D= b² - 4ac = 1 + 8 = 9
T1 = 1/2 и Т2 = -1
Значит 1) cosx = 1/2 ⇒ x = pi/3 + 2pi и x = -pi/3 + 2pi
2) cosx = -1⇒ X= pi+ 2pi