Complex trinomial factoring $ 2 \cos x - 2 = \sin^2 x$

Question

$2 \cos x -2 = \sin^2 x$

I have been trying to solve this equation for the interval $0 \le x \le 2\pi$ .

I figured I should keep them as one, so I put

$2 \cos x -2 = 1 - \cos^{2} x$

however I don't really know how to proceed from there, and i'm pretty sure i'm not actually solving it correctly. I don't know what i'm doing wrong.

Answer 1

HINT:

$$2\cos(x)-2=\sin^2(x)\Longleftrightarrow$$ $$-2+2\cos(x)-\sin^2(x)=0\Longleftrightarrow$$ $$-3+2\cos(x)+\cos^2(x)=0\Longleftrightarrow$$ $$(\cos(x)-1)(3+\cos(x))=0\Longleftrightarrow$$ $$\cos(x)-1=0\Longleftrightarrow \space\space\vee\space\space 3+\cos(x)=0\Longleftrightarrow$$ $$\cos(x)=1\Longleftrightarrow \space\space\vee\space\space \cos(x)=-3\Longleftrightarrow$$ $$x=2\pi n_1 \space\space\vee\space\space x=\pm\cos^{-1}(-3)+2\pi n_2$$

With $n_1,n_2\in\mathbb{Z}$

Answer 2

Hint: collect terms and treat the result as a quadratic equation where the variable is $\cos x$.

Answer 3

$$2 \cos x -2 = \sin^2x$$ $$\Rightarrow 2 \cos x -2 = \frac{1-\cos(2x)}{2}$$ $$\Rightarrow 4 \cos x -4 = 1-\cos(2x)$$ $$\Rightarrow 4 \cos x +\cos(2x) = 5$$ We then note that the $\cos x$ has a range of $[0,1]$, and thus our equation is only equal to $5$ when $\cos(2x) = 1$ and $\cos x = 1$. These both happen at interval of $2\pi$ (think of the unit circle here... only when it is $0$ degrees), and so our answer is $x = 2\pi n$, where $n$ is any integer. We have restricted our domain though from $0$ to $2\pi$, so the answer is $x=0,2\pi$

Answer 4

HINT:

For $$ 2 \cos x - 2 = \sin^2 x$$

Note that $$\sin^2 x = {1-\cos^2x}$$ $$\implies2 \cos x - 2 = {1-\cos^2x}$$

Now solve the quadratic in disguise by letting $u=\cos x$