Let theta in R such that 3 sinh (2 theta)=13- | vedclass

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(A) Given the equation: $3 \sinh (2 	heta) = 13 - 3 e^{2 	heta}$ Substitute $\sinh (2 	heta) = \frac{e^{2 	heta} - e^{-2 	heta}}{2}$: $3 \left( \

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$\frac{1}{2} \log 3$

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(A) Given the equation: $3 \sinh (2 	heta) = 13 - 3 e^{2 	heta}$ Substitute $\sinh (2 	heta) = \frac{e^{2 	heta} - e^{-2 	heta}}{2}$: $3 \left( \frac{e^{2 	heta} - e^{-2 	heta}}{2} ight) = 13 - 3 e^{2 	heta}$ Let $x = e^{2 	heta}$. Since $e^{2 	heta} > 0$,we have $x > 0$. $\frac{3}{2} (x - \frac{1}{x}) = 13 - 3x$ $3(x^2 - 1) = 2x(13 - 3x)$ $3x^2 - 3 = 26x - 6x^2$ $9x^2 - 26x - 3 = 0$ Factoring the quadratic: $9x^2 - 27x + x - 3 = 0$ $9x(x - 3) + 1(x - 3) = 0$ $(9x + 1)(x - 3) = 0$ Since $x > 0$,we must have $x = 3$. $e^{2 	heta} = 3$ Taking the natural logarithm on both sides: $2 	heta = \log 3$ $	heta = \frac{1}{2} \log 3$

Let $\theta \in R$ such that $3 \sinh (2 \theta)=13-3 e^{2 \theta}$,then $\theta=$

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