Solve: $\tan2x=1$ Are there any errors in my work? Thanks in advance! (Sorry for the bad format. I'm still new to this)
$\tan2x=1$
$\frac{2\tan x}{1-\tan^2x}=1$
$2\tan x=1-\tan^2x$
$0=1-\tan^2x-2\tan x$
$0 =-\tan^2x-2\tan x +1$ 
$0=\tan^2x+2\tan x-1$
$\frac{-(2)\sqrt{2^2-4(1)(-1)}}{2(1)}$
$x=0.4142, x=2.4142$ 
$\tan^{-1}(0.4142)$
$x=22.5, x=202.5$
 A: Simple solution:$$\tan 2x=1$$ $$2x=\tan^{-1}(1)$$  $$2x=\frac{\pi}{4}+n\pi\ \text{where n is an integer}$$$$x=\frac{\pi}{8}+\frac{n\pi}{2}=22.5^{\circ}+n\cdot90^{\circ}\ \text{for }n\in\mathbb Z$$
A: That's a very complicated answer!
The obvious answer (using degrees, and using the common fact that $\tan 45 = 1$) is:
$$\tan 2x = 1$$
$$\tan 2x = \tan 45$$
$$2x = 45$$
$$x = 22.5$$
The only thing about going from line 2 to line 3 is that it doesn't guarantee a unique answer - there are other angles that have a tangent of 1.
A: Your solution is not very rigorous after $0 = tan^2x + 2tanx - 1$. After that line, the 
equation should be $tanx = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)}$. 
Also, after that, it's $tanx = \sqrt{2} - 1 \approx 0.4142$ or $tanx = \sqrt{2} + 1 \approx 2.4142$.
Then, $x = tan^{-1}(0.4142) \approx 22.5, 202.5$
A: $$
\begin{align}
\tan 2x&=1\\
\tan 2x&=\tan(180^\circ n+45^\circ)\quad\Rightarrow\quad n\in\mathbb{Z}\\
2x&=180^\circ n+45^\circ\\
x&=90^\circ n+22.5^\circ.
\end{align}
$$
A: 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.
