1. 1-cos (π+x) - sin () = 0

2. 2cos^2 (2π+x) = 3cos () + 2

3. cos2x-cosx=cos3x

1

1+cosx+cos (x/2) = 0

2cos² (x/2) + cos (x/2) = 0

cos (x/2) * (2cos (x/2) + 1) = 0

cos (x/2) = 0⇒x/2=π/2+πk⇒x=π+2πk, k∈z

cos (x/2) = - 1/2

x/2=+-2π/3+2πn, n∈z⇒x=+-4π/3+4πn, n∈z

2

2cos²x-3sinx-2=0

2 (cos²x-1) - 3sinx=0

-2sin²x-3sinx=0

-sinx (2sinx+3) = 0

sinx=0⇒x=πn, n∈z

sinx=-1,5<-1 нет решения

3

сos2x - (cosx+cos3x) = 0

cos2x-2cos2xcosx=0

cos2x (1-2cosx) = 0

cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2, n∈z

cosx=1/2⇒x=+-π/3+2πk, k∈z