Question

Solve the equation
$2\cos h2x+10\sin h2x=5$
giving your answer in terms of a natural logarithm.

Answer 1

$\cos h2x=\frac{1}{2}(e^{2x}+e^{-2x});\sin h2x=\frac{1}{2}(e^{2x}-e^{-2x})$

So, $e^{2x}+e^{-2x}+5e^{2x}+5e^{-2x}=5$

$6e^{2x}-5-4e^{-2x}=0$

$6e^{4x}-5e^{2x}-4=0$

$(3e^{2x}-4)(2e^{2x}+1)=0$

$e^{2x}=\frac{4}{3}$ or $e^{2x}=-\frac{1}{2}$

The only real solution occurs when $e^{2x}>0$

So, $2x=\ln \frac{4}{3}\Rightarrow x=\frac{1}{2}\ln \frac{4}{3}$