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

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Answer 1 · 2016-10-01T00:10:29

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

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 $n$

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