Solving the equation $3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$ Solve the equation:$$3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$$
To solve it, I tried writing the equation in term of $\cos x$. ( I denote $\cos x$ by $c$):
$$3c+\frac12=c^2(1+2c(5-4c^2))$$
$$3c+\frac12=c^2+10c^3-8c^5$$
$$16c^5-20c^3-2c^2+6c+1=0 \qquad\text{Where $c\in[-1,1]$}$$
I tried $c=\pm1,\pm\frac12,0$, but neither of them satisfied the equation. So I don't know how to find $\cos x$.

$3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$
$ \displaystyle 6\cos x + 1 = (1 + \cos 2x) \left(1 + 2 \cos x (3 - 2 \cos2x)\right)$
$ \displaystyle 6\cos x + 1 = (1 + \cos 2x) (1 + 6 \cos x - 4 \cos x \cos 2x)$
$ \displaystyle 6\cos x + 1 = (1 + \cos 2x) (1 + 4 \cos x - 2 \cos 3x)$
$6\cos x + 1 = 5 \cos x + \cos 2x - \cos 5x + 1$
$\cos 2x = \cos x + \cos 5x$
$\cos 2x = 2 \cos 2x \cos 3x$
$\cos 2x = 0$ is a solution and if $\cos 2x \ne 0$,
$\cos 3x = \frac{1}{2}$ is a solution.
