2sin2(θ)-1=-cos(2θ) (trig identity), so cos(θ)=-cos(2θ).
cos(π-2θ)=cos(π+2θ)=-cos(2θ)=cos(θ), therefore:
π±2θ=θ, and θ=π/3 or -π. But cos(-π)=cos(π)=-1.
The solution appears to be θ=π/3 or π. Let's test it.
sin(π/3)=√3/2, sin2(π/3)=¾, 2sin2(π/3)-1=3/2-1=½; cos(⅔π)=-½, so -cos(⅔π)=½, therefore θ=π/3 is a true solution.
sin(π)=0, cos(2π)=1, so 2sin2(π)-1=-cos(2π)=-1⇒θ=π is also a solution.