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

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

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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.
The equation $$ -\cos 	heta = 2 \sin^2 	heta $$ can be rewritten as $$ (2 \cos 	heta + 1)(\cos 	heta - 1) = 0 $$. This leads to two solutions: 1. $$ \cos 	heta = -\frac{1}{2} $$, which occurs at $$ 	heta = \frac{2\pi}{3} + 2\pi k $$ and $$ 	heta = \frac{4\pi}{3} + 2\pi k $$ for any integer $$ k $$. 2. $$ \cos 	heta = 1 $$, which occurs at $$ 	heta = 2\pi k $$ for any integer $$ k $$. So, the general solutions are: - $$ 	heta = \frac{2\pi}{3} + 2\pi k $$ - $$ 	heta = \frac{4\pi}{3} + 2\pi k $$ - $$ 	heta = 2\pi k $$

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

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