# How to find the general solution of $\tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0$

Find the general solution of the equation.

\begin{eqnarray} \tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0\\ \end{eqnarray}

The answers in my textbook are $n\pi$ and $n\pi +\frac{\pi }{3}$.

Previously, I compute the similar questions by using the following operations :

\begin{eqnarray} \\\tan 3x&=&\cot 5x\\ \\\tan 3x&=&-\tan \left(\frac{\pi }{2}+5x\right)\\ \\\tan 3x&=&\tan \left(-\frac{\pi }{2}-5x\right)\\ \\3x&=&-\frac{\pi }{2}-5x\\ \\8x&=&n\pi -\frac{\pi }{2}\\ \\x&=&\frac{n\pi }{8}-\frac{\pi }{16}\\ \end{eqnarray}

But now there is a 3 in front of the second tan.

What should I do?

I do not know whether my method is correct. If you have any other methods, would you mind telling me the methods?

Update 1 : Found one of the answers, are my operations correct?

• Is your question correct? The question seems likely than $tan(x+\pi/3)=3tan(x-\pi/6)$ – MS.Kim Oct 6 '13 at 12:18
• Thx, the question now is correct. – Casper Oct 6 '13 at 12:24

If we assume that your question is to solve the equation $tan(x+\pi/3)+3tan(x-\pi/6)=0$. if we let $A=x+\pi/3, B=x-\pi/6$, then $A-B=\pi/2$, so we can transform the term $tanA$ to $tan(\pi/2+B)$.

and because $tan(\pi/2+B) = -cotB$, the equation will be $-cotB+3tanB=0$, $tan^2B=1/3$ so the general solution of $B=x-\pi/6=n\pi+\pi/6$ or $n\pi-\pi/6$

so $x=n\pi +\pi/3$ or $n\pi$

The solution is given as follows.

I don't know how to use spoiler for a block of code.

\documentclass[preview,border=12pt]{standalone}
\usepackage{amsmath}
\begin{document}
\abovedisplayskip=0pt\relax
\begin{gather*}
\tan (x+\frac{\pi}{3}) + 3\tan (x-\frac{\pi}{6})=0\\
\tan (x+\frac{\pi}{3}) = 3\tan (\frac{\pi}{6}-x)\\
\frac{\tan x + \tan \frac{\pi}{3}}{1-\tan x \tan \frac{\pi}{3}} = 3\frac{\tan \frac{\pi}{6} -\tan x}{1+\tan \frac{\pi}{6}\tan x}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = 3\frac{\frac{\sqrt 3}{3} -\tan x}{1+\frac{\sqrt 3\tan x}{3}}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ \sqrt 3 -3 \tan x}{1+\frac{\sqrt 3\tan x}{3}}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ \sqrt 3 -3 \tan x}{1+\frac{\sqrt 3\tan x}{3}}\times\frac{3}{3}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ 3\sqrt 3 -9 \tan x}{3+ \sqrt 3\tan x}\\
(1- \sqrt 3\tan x) (3\sqrt 3 -9 \tan x) = (\tan x + \sqrt 3)(3+ \sqrt 3\tan x)\\
3\sqrt 3 -9\tan x -9\tan x +9\sqrt 3\tan^2x = 3\tan x +\sqrt 3\tan^2 x +3\sqrt 3 +3\tan x\\
8\sqrt 3\tan^2 x  -24 \tan x=0\\
\sqrt 3\tan^2 x -3\tan x=0\\
\tan x (\sqrt 3 \tan x -3)=0\\
\tan x =0 \text{ or } \sqrt 3 \tan x -3 =0\\
\tan x=0 \text{ or } \tan x =\sqrt 3\\
x=n\pi \text{ or } x=\frac{\pi}{3}+n\pi
\end{gather*}

\end{document}

• =口= This operation must take so much time when during exam...Is it the best way? – Casper Oct 6 '13 at 13:42
• @CasperLi: Not much even though you use LaTeX during the exam. :-) – kiss my armpit Oct 6 '13 at 13:44
• Are there any fast methods to input math/LaTeX? ~.~ – Casper Oct 6 '13 at 13:53
• @CasperLi: It depends on the text editor you are using. If there is a shortcut key for each macro then you can type much faster. But I think the important point is your typing speed. :-) – kiss my armpit Oct 6 '13 at 13:55