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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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( \

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $\theta=$

Similar Questions

If ${a^x} = bc$,${b^y} = ca$,and ${c^z} = ab$,then find the value of $xyz$.

The number of solutions of $\log_{2}(x + 5) = 6 - x$ is

The number of solutions of the equation $\log _{(x+1)}(2 x^{2}+7 x+5)+\log _{(2 x+5)}(x+1)^{2}-4=0$ for $x > 0$ is:

The set of real values of $x$ that satisfy the inequality ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$ is:

The number of integral solutions $x$ of $\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^2 \geq 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module