How do you find the general solutions for $\cos (2x) + 5\cos (x) + 3 = 0$?

Hint: First we will evaluate the right-hand of equation and then further the left-hand side of the equation. We will use the identity ${\sin ^2}x + {\cos ^2}x = 1$. Then we will try to factorise and simplify the terms so that the left-hand side matches the right-hand side.

Complete step-by-step solution:
We will start solving this question by directly using the double angle formula for $\cos $.
$
\Rightarrow \cos (2x) + 5\cos (x) + 3 = 0 \\
\Rightarrow 2{\cos ^2}(x) + 5\cos (x) + 3 = 0 \\
 $
Now, if we factorise the equation further we will get,
$
\Rightarrow 2{\cos ^2}(x) + 5\cos (x) + 3 = 0 \\
\Rightarrow (2\cos (x) + 3)(\cos (x) + 1) = 0 \\
 $
Now if we solve separately for the value of $\cos $ we will get,
$
\Rightarrow (2\cos (x) + 3)(\cos (x) + 1) = 0 \\
\Rightarrow 2\cos (x) + 3 = 0 \\
\Rightarrow 2\cos (x) = - 3 \\
\Rightarrow \cos (x) = \dfrac{{ - 3}}{2} \\
 $
And the other value will be given by,
$
\Rightarrow (2\cos (x) + 3)(\cos (x) + 1) = 0 \\
\Rightarrow \cos (x) + 1 = 0 \\
\Rightarrow \cos (x) = - 1 \\
 $
Hence, the values of $\cos $ will be $ - 1,\dfrac{{ - 3}}{2}$.
Now, we know that $\cos (x) \in [ - 1,1]$ for all values of $x$.
And here, the value $\cos x = \dfrac{{ - 3}}{2}$ is not included in the range and the other value which is $\cos x = - 1$ is within the range which also implies the same way $\,\,\,\,\,x = 3\pi $ within the range $[0,2\pi ]$.
And if the range is not restricted, the value will be $x = \pi + 2n\pi ,\forall n \in Z$

Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities ${\sin ^2}x + {\cos ^2}x = 1$ and $\cos 2x = 2{\cos ^2}x - 1$. While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity. Also, remember that the range of $\cos $ function is from $ - 1\,$ to $ + 1$ and the range of $\sin $ function is also from $ - 1\,$ to $ + 1$.