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


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

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

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

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

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


$x=0.4142, x=2.4142$


$x=22.5, x=202.5$


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

  • $\begingroup$ So it would it be 25.5 and 112.5 for one interval? $\endgroup$ – Learner May 7 '14 at 4:40
  • 3
    $\begingroup$ Yes. Notice that there are only $2$ values that satisfy this equation on $[0^{\circ},360^{\circ}]$. In particular, when $n=0,1$. $\endgroup$ – homegrown May 7 '14 at 4:49
  • $\begingroup$ @jnh: I think you mean there are only $2$ values of $2x$ on $[0^{\circ},360^{\circ}]$ (and therefore $2$ values of $x$ on $[0^{\circ},180^{\circ}]$). $\endgroup$ – TonyK Jun 19 '14 at 10:00

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.


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


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

  • $\begingroup$ $0<=x<360$ .............. $\endgroup$ – Learner May 7 '14 at 3:16
  • $\begingroup$ I'm not sure how you got the third one: 247.5 $\endgroup$ – Learner May 7 '14 at 3:24
  • $\begingroup$ $67.5^{\circ}$ and $247.5^{\circ}$ are not solutions to this. Note that $\tan(2\cdot 67.5^{\circ})=\tan(2\cdot 247.5^{\circ})=-1$ $\endgroup$ – homegrown May 7 '14 at 3:31

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.

  • $\begingroup$ This is the way we're taught. I'm not looking for efficiency, just that the method I'm forced to use was done correctly. $\endgroup$ – Learner May 7 '14 at 3:28
  • $\begingroup$ Well, if your instructor insisted that you do this by calling in the double-angle formula, then I would replace my second paragraph with a criticism of the instructor for making things unnecessarily difficult. Indeed, if the question had been to solve $\tan(9x/2)=1$, it would have been frustratingly difficult to use the nine-fold angle formula and the half-angle formulas, whereas @user135348’s method would have worked fine. $\endgroup$ – Lubin May 7 '14 at 3:32

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