How to find the solution set for $\tan ^2 x= (\sqrt{2}-1)^2$?

Question

The problem is as follows:

Find the set solution for the equation from below:

$\sqrt{2}-\tan (2x)=\cot \left(\frac{\pi}{4}+x\right)$

Well after doing all the necessary algebra I'm getting the following expression from below.

$\tan ^2 x= (\sqrt{2}-1)^2$

Well my intuition tells me that I have to take the inverse function for both sides of the equation not before taking the square rooth in both sides which will yield two solutions:

$\tan x= \pm \sqrt{2}-1$

But taking the inverse function will yield:

$x=\tan^{-1}\left(\sqrt{2}-1\right)$

But what it is in the brackets it is not an arc which I can remember. Could it be that I got something wrong here or what?.

Can someone help me here please?. I will really appreciate to know how exactly can I get the set solution for this.

Answer 1

First, it should be $$\tan x = \pm \color{red}{(}\sqrt{2} - 1 \color{red}{)}.$$ Second, I think a better approach is to write instead $$\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{(1 - \cos 2x)/2}{(1 + \cos 2x)/2} = \frac{1 - \cos 2x}{1 + \cos 2x} = (\sqrt{2} - 1)^2.$$ Solving this equation for $\cos 2x$ yields $$\cos 2x = \frac{1}{\sqrt{2}},$$ hence $$2x = \frac{\pi}{4} + 2\pi k, \quad 2x = -\frac{\pi}{4} + 2\pi k, \quad k \in \mathbb Z,$$ giving $$x = \pm \frac{\pi}{8} + \pi k, \quad k \in \mathbb Z.$$ We can then verify this solution by direct substitution, noting that $\tan x$ is periodic with period $\pi$: $$\tan^2 \left(\pm \frac{\pi}{8}\right) = 3 - 2\sqrt{2}.$$

Answer 2

$$\sqrt2-1=\csc\pi/4-\cot\pi/4=\cdots=\tan\pi/8$$

Now if $\tan^2x=\tan^2y\implies\sin^2x=\cdots=\sin^2y$

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

$$x=n\pi\pm y$$