Question

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

Answer 1

`x = -pi/4 + npi" "` for any integer `n`

Explanation:

Given:

`arccos(tan x) = 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(-1) + npi" "` for any integer `n`

By considering a right angled triangle with sides `1, 1, sqrt(2)` and angles `pi/4, pi/4, pi/2` we can deduce that `arctan(1) = pi/4`

Then `tan x` is an odd function, so `arctan(-1) = -pi/4`

So the solutions of our original equation are:

`x = -pi/4 + npi" "` for any integer `n`