Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!

One can use trigonometric functions to avoid squaring both sides to get a degree 4 polynomial. Clearly $x>1$. Let $x=\sec t$, $t\in(0,\frac{\pi}{2})$. Then the equation becomes
$$ \sec t+\csc t=\frac{35}{12} $$
or
$$ \frac{\sin t+\cos t}{\sin t\cos t}=\frac{35}{12}. $$
Squaring both sides gives
$$ \frac{1+\sin(2t)}{\sin^2(2t)}=\frac{35^2}{24^2} $$
which is equivalent to
$$ 35^2\sin^2(2t)-24^2\sin(2t)-24^2=0 $$
or
$$ (25\sin(2t)-24)(49\sin(2t)+24)=0. $$
Since $\sin(2t)>0$, one has
$$ \sin(2t)=\frac{24}{25} $$
which gives
$$ \cos(2t)=\pm\frac{7}{25}. $$
So
$$ \cos t=\sqrt{\frac{1+\cos(2t)}{2}}=\frac{3}{5} \text{ or } \frac{4}{5} $$
and hence
$$ x=\frac{5}{3} \text{ or }\frac{5}{4}. $$
