Exponential equation with $\cos(x)$ both in base and exponent

I'm trying to solve the following equation $|\cos(x)|^{2\cos(x)+1} = 1$, but none of the standard methods I know work, because the variable is both in base and exponent. What shall I do?

As base is positive , You can use logarithms :

$|\cos(x)|^{2\cos(x)+1} = 1$

$\ln |\cos(x)|^{2\cos(x)+1} = \ln 1$

$({2\cos(x)+1}) \times \ln|\cos(x)| = 0$

Either

${2\cos(x)+1}=0$

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

$\implies x = 2n\pi \pm \frac {2\pi}{3} , n\in\mathbb Z$

Or

$\ln|\cos(x)|=0 \implies |\cos(x)|=1$

$\implies\cos(x)=\pm1\implies x=n \pi ,n\in\mathbb Z$

So, Finally

$x = \bigcup_{n\in\mathbb Z}$ {$n\pi,2n\pi \pm \frac {2\pi}{3}$}

For any problem of the form $a^b=1$, we have three cases:

$$a=1,b\in\mathbb R\\b=0,a\in\mathbb R\setminus \{0\}\\a=-1,b/2\in\mathbb N$$

Thus, it is easy enough to see that $a=-1$ is not possible due to the absolue value bars, so we are left with

$$a=|\cos(x)|=1\implies\cos(x)=\pm1\implies x=\pi n,n\in\mathbb Z$$

$$b=2\cos(x)+1=0\implies\cos(x)=-\frac12\implies x=\pm\frac{2\pi}3+2\pi n,n\in\mathbb Z$$