Solve 2cos2θ+cosθ−1=0
2nπ ∀ n∈N
2nπ∓π3 ∀ n∈N
2nπ∓π6 ∀ n∈N
(2n+1)π6 ∀ n∈N
⇒cosθ=−1∓√1+84 cosθ=12 or cosθ=−1
We know that the general solution forumla for the equation cosx=cosα is x=2nπ±α ⇒θ=2nπ±π3 or π=(2n±1)π