Solve: $\tan2x=1$

Question

Are there any errors in my work? Thanks in advance! (Sorry for the bad format. I'm still new to this)

$\tan2x=1$

$\frac{2\tan x}{1-\tan^2x}=1$

$2\tan x=1-\tan^2x$

$0=1-\tan^2x-2\tan x$

$0 =-\tan^2x-2\tan x +1$

$0=\tan^2x+2\tan x-1$

$\frac{-(2)\sqrt{2^2-4(1)(-1)}}{2(1)}$

$x=0.4142, x=2.4142$

$\tan^{-1}(0.4142)$

$x=22.5, x=202.5$

Answer 1

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, $x = tan^{-1}(0.4142) \approx 22.5, 202.5$

Answer 2

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.

Answer 3

Simple solution:$$\tan 2x=1$$ $$2x=\tan^{-1}(1)$$ $$2x=\frac{\pi}{4}+n\pi\ \text{where n is an integer}$$$$x=\frac{\pi}{8}+\frac{n\pi}{2}=22.5^{\circ}+n\cdot90^{\circ}\ \text{for }n\in\mathbb Z$$

Answer 4

$$ \begin{align} \tan 2x&=1\\ \tan 2x&=\tan(180^\circ n+45^\circ)\quad\Rightarrow\quad n\in\mathbb{Z}\\ 2x&=180^\circ n+45^\circ\\ x&=90^\circ n+22.5^\circ. \end{align} $$

Answer 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.