If
Question:

If $\frac{\tan 3 x-1}{\tan 3 x+1}=\sqrt{3}$, then $x=$

Solution:

$\frac{\tan 3 x-1}{\tan 3 x+1}=\sqrt{3}$

i. e $\tan 3 x-1=\sqrt{3} \tan 3 x+\sqrt{3}$

$\tan 3 x-1=\sqrt{3} \tan 3 x+\sqrt{3}$

i.e $-1-\sqrt{3}=\tan 3 x(\sqrt{3}-1)$

i. e $\tan 3 x=-\frac{(1+\sqrt{3})}{\sqrt{3}-1}$

$\tan 3 x=-\tan 75^{\circ}$

$=\tan \left(180-75^{\circ}\right) \quad[\because \tan (180-\theta)=-\tan \theta]$

$\tan 3 x=\tan \frac{7 \pi}{12}$

i. e $3 x=n \pi+\frac{7 \pi}{12}$

i. e $x=\frac{n \pi}{3}+\frac{7 \pi}{36}$ is the general solution

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