Solve the equation on the interval [0,2pi) 2cos² θ/2 -3cosθ=0

To solve this problem, you first have to know the half angle identify of cosine.
cos(θ/2)=+/-√(1+cosθ)/2, so  cos² (θ/2)=(1+cosθ)/2, and 2cos² (θ/2)=1+cosθ

2cos² (θ/2) -3cosθ=0
1+cosθ-3cosθ=0
2cosθ=1
cosθ=1/2
θ=π/3, or (5/3)π

You didn’t put the necessary image to solve the question : p