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If the general solution of the equation $\frac{\tan 3 x-1}{\tan 3 x+1}=\sqrt{3}$ is $x=\frac{\mathrm{n} \pi}{\mathrm{p}}+\frac{7 \pi}{\mathrm{q}}, \mathrm{n}, \mathrm{p}, \mathrm{q}, \in \mathrm{Z}$, then $\frac{p}{q}$ is
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Verified Answer
The correct answer is:
$\frac{1}{12}$
$\begin{aligned} & \frac{\tan 3 x-1}{\tan 3 x+1}=\sqrt{3} \\ \therefore \quad & \frac{\tan 3 x-\tan \frac{\pi}{4}}{1+(\tan 3 x)\left(\tan \frac{\pi}{4}\right)}=\sqrt{3} \\ \therefore \quad & \tan \left(3 x-\frac{\pi}{4}\right)=\tan \left(\frac{\pi}{3}\right) \\ \therefore \quad & 3 x-\frac{\pi}{4}=n \pi+\frac{\pi}{3}\end{aligned}$
$\begin{aligned}
& \therefore \quad 3 x=\mathrm{n} \pi+\frac{\pi}{3}+\frac{\pi}{4} \\
& \therefore \quad 3 x=\mathrm{n} \pi+\frac{7 \pi}{12} \\
& \therefore \quad x=\frac{\mathrm{n} \pi}{3}+\frac{7 \pi}{36} \\
&
\end{aligned}$
Comparing with $\frac{\mathrm{n} \pi}{\mathrm{p}}+\frac{7 \pi}{\mathrm{q}}$, we get
$\begin{aligned}
& \mathrm{p}=3, \mathrm{q}=36 \\
\therefore \quad & \frac{\mathrm{p}}{\mathrm{q}}=\frac{3}{36}=\frac{1}{12}
\end{aligned}$
$\begin{aligned}
& \therefore \quad 3 x=\mathrm{n} \pi+\frac{\pi}{3}+\frac{\pi}{4} \\
& \therefore \quad 3 x=\mathrm{n} \pi+\frac{7 \pi}{12} \\
& \therefore \quad x=\frac{\mathrm{n} \pi}{3}+\frac{7 \pi}{36} \\
&
\end{aligned}$
Comparing with $\frac{\mathrm{n} \pi}{\mathrm{p}}+\frac{7 \pi}{\mathrm{q}}$, we get
$\begin{aligned}
& \mathrm{p}=3, \mathrm{q}=36 \\
\therefore \quad & \frac{\mathrm{p}}{\mathrm{q}}=\frac{3}{36}=\frac{1}{12}
\end{aligned}$
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