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Let $\theta \in R$ such that $3 \sinh (2 \theta)=13-3 e^{2 \theta}$, then $\theta=$
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Verified Answer
The correct answer is:
$\frac{1}{2} \log 3$
$$
\begin{aligned}
& \text { } 3 \sinh (2 \theta)=13-3 e^{2 \theta} \\
& 3\left(\frac{e^{2 \theta}-e^{-2 \theta}}{2}\right)=13-3 e^{2 \theta}
\end{aligned}
$$
Let $x=e^{2 \theta}$
$\begin{aligned} & \frac{3}{2}\left(x-\frac{1}{x}\right)=13-3 x \\ \Rightarrow & 3 x^2-3=2 x(13-3 x) \Rightarrow 3 x^2-3=26 x-6 x^2 \\ \Rightarrow & 9 x^2-26 x-3=0 \Rightarrow 9 x^2-27 x+x-3=0 \\ \Rightarrow & 9 x(x-3)+1(x-3)=0 \Rightarrow(9 x+1)(x-3)=0 \\ \Rightarrow & x=3 \left[\because x \neq-\frac{1}{9}\right]
\\ \therefore & e^{2 \theta}=3 \\ \Rightarrow & 2 \theta=\log 3 \Rightarrow \theta=\frac{1}{2} \log 3\end{aligned}$
\begin{aligned}
& \text { } 3 \sinh (2 \theta)=13-3 e^{2 \theta} \\
& 3\left(\frac{e^{2 \theta}-e^{-2 \theta}}{2}\right)=13-3 e^{2 \theta}
\end{aligned}
$$
Let $x=e^{2 \theta}$
$\begin{aligned} & \frac{3}{2}\left(x-\frac{1}{x}\right)=13-3 x \\ \Rightarrow & 3 x^2-3=2 x(13-3 x) \Rightarrow 3 x^2-3=26 x-6 x^2 \\ \Rightarrow & 9 x^2-26 x-3=0 \Rightarrow 9 x^2-27 x+x-3=0 \\ \Rightarrow & 9 x(x-3)+1(x-3)=0 \Rightarrow(9 x+1)(x-3)=0 \\ \Rightarrow & x=3 \left[\because x \neq-\frac{1}{9}\right]
\\ \therefore & e^{2 \theta}=3 \\ \Rightarrow & 2 \theta=\log 3 \Rightarrow \theta=\frac{1}{2} \log 3\end{aligned}$
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