Solve the given trigonometry equation

[chwala]

Homework Statement

Find the exact value given the equation;

$\sinh^{-1} x = 2\cosh^{-1} (2)$

Relevant Equations

Trigonometry

In my approach i have the following lines

$\ln (x + \sqrt{x^2+1}) = 2\ln (2+\sqrt 3)$

$\ln (x + \sqrt{x^2+1} = \ln (2+\sqrt 3)^2$

$⇒x+ \sqrt{x^2+1} =(2+\sqrt 3)^2$

$\sqrt{x^2+1}=-x +7+4\sqrt{3}$

$x^2+1 = x^2-14x-8\sqrt 3 x + 56\sqrt 3 +97$

$1 = -14x-8\sqrt 3 x + 56\sqrt 3 +97$

$14x+8\sqrt 3 x = 96+56\sqrt 3$

$(14+8\sqrt 3)x = 96+56\sqrt 3$

$x= \dfrac{96+56\sqrt 3}{14+8\sqrt 3} = \dfrac{1344-768\sqrt 3 +784\sqrt 3-1344}{196-192}= \dfrac{16\sqrt 3}{4}=4\sqrt 3$

Bingo ...any insight or alternative approach is welcome.

 

[PeroK]

Using the double angle formula looks easier:
$$x = \sinh(2\cosh^{-1}(2)) = 4\sinh(\cosh^{-1}(2)) = 4\sqrt{\cosh^2(\cosh^{-1}(2)) - 1} = 4\sqrt 3$$

 

[Orodruin]

Apply sinh:
$$
x = \sinh(2\cosh^{-1}(2))
= 2\sinh(\cosh^{-1}(2)) \cosh(\cosh^{-1}(2))
= 4\sinh(\cosh^{-1}(2))
$$
Since $\sinh = \sqrt{\cosh^2-1}$
$$
\sinh(\cosh^{-1}(2)) = \sqrt{4-1} = \sqrt 3
$$
anf therefore
$$
x = 4\sqrt 3
$$

Edit: Cross posted with @PeroK - at least we said exactly the same …