If one solution of the equation cosh x - frac | vedclass

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

Question

(D) Given equation: $\cosh x - \frac{4}{5} \sinh x = 1$ Multiply by $5$: $5 \cosh x - 4 \sinh x = 5$ Substitute $\cosh x = \frac{e^x + e^{-x}}{2}$ and

Vedclass · Accepted Answer

$2 \log 3$

Vedclass · Answer

(D) Given equation: $\cosh x - \frac{4}{5} \sinh x = 1$ Multiply by $5$: $5 \cosh x - 4 \sinh x = 5$ Substitute $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$: $5 \left(\frac{e^x + e^{-x}}{2}\right) - 4 \left(\frac{e^x - e^{-x}}{2}\right) = 5$ Multiply by $2$: $5(e^x + e^{-x}) - 4(e^x - e^{-x}) = 10$ $5e^x + 5e^{-x} - 4e^x + 4e^{-x} = 10$ $e^x + 9e^{-x} = 10$ Multiply by $e^x$: $(e^x)^2 - 10e^x + 9 = 0$ $(e^x - 1)(e^x - 9) = 0$ Case $1$: $e^x = 1 \Rightarrow x = 0$ Case $2$: $e^x = 9 \Rightarrow x = \ln 9 = \ln(3^2) = 2 \ln 3$ Thus,the other solution is $x = 2 \log 3$.

If one solution of the equation $\cosh x - \frac{4}{5} \sinh x = 1$ is $x = 0$,then the other solution is $x =$

Similar Questions

If $0 \leq x \leq 3$ and $0 \leq y \leq 3$,then the number of solutions $(x, y)$ of the equation $\left(\sqrt{\sin^2 x - \sin x + \frac{1}{2}}\right) 2^{\sec^2 y} = 1$ is

If $\cosh \beta = \sec \alpha \cos \theta$ and $\sinh \beta = \operatorname{cosec} \alpha \sin \theta$,then $\sinh^2 \beta =$

The value of $\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$ is

If $\sin x + \sin y = \alpha$ and $\cos x + \cos y = \beta$,then $\operatorname{cosec}(x + y) = $

If $\sinh x = -\frac{1}{2}$,then $\tanh 2x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module