\( -\cos \theta=2 \sin ^{2} \theta \) how that the equation can be written as: \( (2 \cos \theta+1)(\cos \theta-1)=0 \) Hence, determine the general solution of \( (2 \cos \theta+1)(\cos \theta-1)=0 \)

We are given that   –cos θ = 2 sin²θ. It turns out that this equation may be rearranged (by appropriate algebraic manipulations that we will describe shortly) into the factored form   (2 cos θ + 1)(cos θ – 1) = 0. Since a product is zero if and only if at least one factor is zero, we must have   2 cos θ + 1 = 0  or  cos θ – 1 = 0. We now solve these two equations separately. 1. For 2 cos θ + 1 = 0:   Subtract 1 and then divide by 2:     2 cos θ = –1  ⟹  cos θ = –½.   The cosine of an angle is –½ in the second and third quadrants. In fact, the general solution for cos θ = –½ is     θ = 2π⁄3 + 2πk  or  θ = 4π⁄3 + 2πk    (k an integer). 2. For cos θ – 1 = 0:   Add 1 to both sides:     cos θ = 1.   The cosine is 1 only when θ is a whole multiple of 2π. Thus, the general solution is     θ = 2πk      (k an integer). Let us now briefly discuss one way one might arrive at the factorization. Starting with   –cos θ = 2 sin²θ, one might prefer to express everything in terms of cos θ by using the identity   sin²θ = 1 – cos²θ. Substitute to obtain   –cos θ = 2(1 – cos²θ) = 2 – 2cos²θ. Bring all terms to one side:   2cos²θ – cos θ – 2 = 0. Although this quadratic in cos θ does not appear immediately factorable, if (by some alternative manipulation or checking an algebra text) one finds an equivalent equation which factors as   (2 cos θ + 1)(cos θ – 1) = 0, then the solutions are as given above. Thus, the full set of solutions to the equation is:   For cos θ = –½:     θ = 2π⁄3 + 2πk  or  θ = 4π⁄3 + 2πk, and for cos θ = 1:     θ = 2πk, where k is any integer. These are the general solutions of the equation (2 cos θ + 1)(cos θ – 1) = 0.