how to solve equation with cos I have this equation $\cos2x +5 \cos x + 3=0$. To solve it I rewrite $\cos2x$ to $2 \cos^{2} x- 1$ and set $\cos = t$.
I get the following equation $2t^2 - 1 +5t +3 = 0$ with that and then divide the equation with two $t^2 +\frac{5}{2} t +1 = 0$. I solve this equation and get two $t$, $t_1 = -2$ and $t_2 = - \frac {1}{2}$. $t_2$ is the valid because $t$ can't be larger than 1.
From here on I don't know how to use $t$ to solve this equation $\cos2x +5 \cos x + 3=0$.
Can anyone explain what to do next and how to solve this equation?
Thanks!!

It is not necessary to use substitutions when what you're working with isn't going to be too long:
$$
\cos 2x +5 \cos x + 3 = 0\\
\implies 2\cos^2 x - 1 + 5\cos x + 3 = 0\\
\implies 2\cos^2 x + 5\cos x + 2 = 0\\
\implies \cos^2 x + \frac{5}{2}\cos x + 1 = 0\\
\implies \cos^2 x + \left(2 + \frac{1}{2}\right)\cos x + 2\cdot \frac{1}{2} = 0\\
\implies (\cos x + 2)(\cos x + \frac{1}{2}) = 0
$$
Now, $\cos x \neq -2$ since $\cos x \in \left[-1,1\right]$. So we can reject that.
$$\therefore \space\cos x = - \frac{1}{2} \\
\implies \cos x = -\sin{\frac{\pi}{6}} = \cos\left(\frac{\pi}{2} + \frac{\pi}{6}\right) = \cos \left(\frac{2\pi}{3}\right)\\
\implies x = 2n\pi \pm \frac{2\pi}{3} , \space n\in \mathbb Z\\
\implies x = 2\pi \left(n \pm \frac{2}{3}\right), \space n\in \mathbb Z $$

                   


Trivial yet Important Facts used:


*

*$x^2 + \left(\alpha + \beta\right)x + \alpha\beta = 0 \iff (x+\alpha)(x+\beta) = 0$

*$\cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta$

*$\cos x = \cos \theta \implies x = 2n\pi \pm \theta$

