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?
Thank you for your attention.
Update 1 : Found one of the answers, are my operations correct?
Best Answer
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 $n\pi$