<s>[INST] 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!!
 [/INST] 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:


*

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

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

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

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