Найдите критические точки функции

1) y=0.5 sin2x+sinx

2) y=0.5 cos2x-cosx

3) y=0.5 cos2x+cosx

1) y'=cos2x+cosx=0

2cos²x-1+cosx=0

cosx=a

2a²+a-1=0

D=1+8=9

a1 = (-1-3) / 4=-1⇒cosx=-1⇒x=π+2πn

a2 = (-1+3) / 4=1/2⇒cosx=1/2⇒x=+-π/3+2πn

2) y'=-sin2x+sinx=0

-2sinxcosx+sinx=0

-sinx (2cosx-1) = 0

sinx=0⇒x=πn

cosx=1/2⇒x=+-π/3+2πn

3) y'=-sin2x-sinx=0

-2sinxcosx-sinx=0

-sinx (2cosx+1) = 0

sinx=0⇒x=πn

cosx=-1/2⇒x=+-2π/3+2πn