Question

Solve the equation $5\sin hx+\cos hx+5=0$ giving your answer in the form of $\ln a$ where $a$ is a rational number.

Answer 1

$5\sinh x+\cosh x+5=0$

$5\frac{e^x-e^{-x}}{2}+\frac{e^x+e^{-x}}{2}+5=0$

Let $e^x=t$

$\frac{5}{2}(t-\frac{1}{t})+\frac{1}{2}(t+\frac{1}{t})+5=0$

$\Rightarrow \frac{5}{2}t+\frac{1}{2}t+5-\frac{5}{2t}+\frac{1}{2t}=0$

$\Rightarrow 3t+5-\frac{2}{t}=0$

$\Rightarrow \frac{3t^2+5t-2}{t}=0$

$\Rightarrow 3t^2+5t-2=0$

$t=\frac{-5\pm \sqrt{25+24}}{2\times 3}$

$=\frac{-5\pm \sqrt{49}}{6}$

$e^x=\frac{-5+7}{6}$ $(\because e^x>0(always\left)\right)$

$x=\ln \left(\frac{1}{3}\right)$