what is 2cos^2(θ)−1

I assume you want to solve this:

2cos^2(θ)−1 = 0

Add 1 to both sides

2cos^2(θ) =1

Divide by 2 on both sides

cos^2(θ) = 1/2

Take the square root of both sides

cos(θ)= ± 1/√2

So θ  = pi/4, 3pi/4, 5pi/4 and 7pi/4  in the interval  [0, 2pi]

what is 2cos^2(θ)−1    ?

$$\cos{(2\theta)}=\cos^2{(\theta)} -\underbrace{ \sin^2{(\theta)}}_{=( 1- \cos^2{(\theta)} ) }\\ \cos{(2\theta)}=\cos^2{(\theta)} -1 +\cos^2{(\theta)} \\ \cos{(2\theta)} =2\cos^2{(\theta)} -1 \\\\ \boxed{ 2\cos^2{(\theta)} -1 = \cos{(2\theta)} }$$