First observe 2 cos(2 θ) = 2 − cos (2 θ)=>3cos(2t)=2

=>cos(2t)=2/3.

From cos(x+y)=cos(x)cos(y)-sin(x)sin(y) and sin^2+cos^2=1,

cos(2t)=cos^2(t)-sin^2(t)=1-2sin^2(t).

Hence 1-2sin^2(t)=2/3=> 1-2/3=2sin^2(t)

=>sin^2(t)=1/6=> sin(t)=+/-1/sqrt(6).

So let x=arcsin[1/sqrt(6)].

Then the solutions are

x and 2*Pi-x.

Hope that helps.