Solve: $\tan2x=1$ Are there any errors in my work? Thanks in advance! (Sorry for the bad format. I'm still new to this)
$\tan2x=1$
$\frac{2\tan x}{1-\tan^2x}=1$
$2\tan x=1-\tan^2x$
$0=1-\tan^2x-2\tan x$
$0 =-\tan^2x-2\tan x +1$ 
$0=\tan^2x+2\tan x-1$
$\frac{-(2)\sqrt{2^2-4(1)(-1)}}{2(1)}$
$x=0.4142, x=2.4142$ 
$\tan^{-1}(0.4142)$
$x=22.5, x=202.5$

This is an expansion of @user135348’s perfect answer. The question is talking about the angle $2x$, whose tangent is known to be $1$. But what are the angles with tangent unity? They are just $45^\circ + k\cdot180^\circ$ for all integer values of $k$. Since $2x=45^\circ+k\cdot180^\circ$, you get $x=22.5^\circ+k\cdot90^\circ$.
What happened to you is something that happens to all of us from time to time: you got seduced into thinking that the problem was hard, when in fact it is very easy.
