If 4 + 6 e 2 x + 1 tan h x = 11 cos h x + 11 sin h x , then x =

Given:

$4+6\left(e^{2x}+1\right)\tan hx=11\cos hx+11\sin hx$

$\Rightarrow 4+6\left(e^{2x}+1\right)\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)=11\left(\frac{e^x+e^{-x}}{2}\right)+11\left(\frac{e^x-e^{-x}}{2}\right)$

$\Rightarrow 4+6\left(e^{2x}+1\right)\left(\frac{e^{2x}-1}{e^{2x}+1}\right)=11e^x$

$\Rightarrow 4+6\left(e^{2x}-1\right)=11e^x$

$\Rightarrow 6e^{2x}-11e^x-2=0$

Put $e^x=t; t>0$

$\Rightarrow 6t^2-11t-2=0$

$\Rightarrow t=\frac{11\pm \sqrt{121+48}}{12}$

$\Rightarrow t=2, -\frac{1}{6}$

Since, $t>0$, so

$t=2$

$\Rightarrow e^x=2$

$\Rightarrow x{\log }_e\left(e\right)={\log }_e\left(2\right)$

$\Rightarrow x={\log }_e\left(2\right)$