2sin^2x-3cosx-2=0

2sin^2x-cosx=0

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

2 * (1-cos^2 (x)) - 3cosx-2=0⇒-2cos^2 (x) - 3cosx=0⇒2cos^2 (x) + 3cosx=0⇒

cosx (2cosx+3) = 0⇒

a) cosx=0⇒x=π/2+πn

б) 2cosx+3=0⇒cosx=-1.5 - решений нет

Ответ: x=π/2+πn

2) 2 * (1-cos^2 (x)) - cosx=0⇒-2cos^2 (x) - cosx+2=0⇒2cos^2 (x) + cosx-2=0⇒

a) cosx = (-1+√1+16) / 4⇒cosx = (-1+√17) / 4⇒

x1=arccos (-1+√17) / 4 и x2=-arccos (-1+√17) / 4

б) cosx = (-1-√1+16) / 4⇒cosx = (-1-√17) / 4< (-1) ⇒решений нет