# How do you solve arccos(tan x)= pi?

Oct 1, 2016

$x = - \frac{\pi}{4} + n \pi \text{ }$ for any integer $n$

#### Explanation:

Given:

$\arccos \left(\tan x\right) = \pi$

Take the cosine of both sides to get:

$\tan x = - 1$

The period of $\tan x$ is $\pi$, so we have solutions:

$x = \arctan \left(- 1\right) + n \pi \text{ }$ for any integer $n$

By considering a right angled triangle with sides $1 , 1 , \sqrt{2}$ and angles $\frac{\pi}{4} , \frac{\pi}{4} , \frac{\pi}{2}$ we can deduce that $\arctan \left(1\right) = \frac{\pi}{4}$

Then $\tan x$ is an odd function, so $\arctan \left(- 1\right) = - \frac{\pi}{4}$

So the solutions of our original equation are:

$x = - \frac{\pi}{4} + n \pi \text{ }$ for any integer $n$