How do you solve $tan x = 1$ $tanx=1$ and find all exact general solutions?

Question

77139 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-02T19:38:52

$x in {((2n+1)pi)/2, n in ZZ}$

If $tan x = 1$ , then $sin x = cos x$ .

We know this is true for $x = pi/4$ as a base case.

Now, the tangent function is periodic with it's period
$p = npi$ , where $n$ is an integer.

So $tan x = tan (x+npi)$ . Go back to the original equality :

$tan x = tan (x+npi) = 1$

The first positive value $x_0$ for which $tan x = 1$ is, as stated before, $pi/4$ .

$tan x_0 = tan(x_0 + npi) =1$

$tan(pi/4 + npi) = tan (((4n+1)pi)/4) = 1$

Which means $tan x = 1$ if and only if $x$ is a number of the form
$((4n+1)pi)/4$ :

$x in {((4n+1)pi)/4 , n in ZZ}$ .

How do you solve tanx=1tanx=1tanx=1 and find all exact general solutions?