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 :

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

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

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



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

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


$\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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.