Question

2Cos^(2)theta+Cos^(2)2 theta-2Cos^(2)theta*Cos2 theta=1​

Answer 1

As Sasha pointed out, one can simply add cos2θ to both sides and solve for the quadratic equation. Here is what occurs:

2cos2θ−cosθ−1+cos2θ=sin2θ+cos2θ

3cos2θ−cosθ−1=1

3cos2θ−cosθ−2=0

Let w=cosθ. Then we get:

3w2−w−2=0

3w2−3w+2w−2=0

3w(w−1)+2(w−1)=0

(3w+2)(w−1)=0

So w=−23 or w=1, which means cosθ=−23 or cosθ=1. This gives us the answers θ=2kπ,k∈Z and θ=±cos−1(−23)+2kπ,k∈Z