Problem source

The real numbers $x$ and $y$ satisfy the simultaneous equations
$$
\sinh (2x) = \cosh y
\qquad\hbox{and}\qquad
\sinh(2y) = 2 \cosh x.
$$
Show that $\sinh^2 y$ is a root of the equation
$$
4t^3 + 4t^2 -4t -1=0
$$
and demonstrate that this gives at most one valid solution for $y$. Show that the relevant value of $t$ lies between $0.7$ and $0.8$, and use an
iterative process to find $t$ to 6 decimal places.
Find $y$ and hence find $x$, checking your answers and stating the final  answers to four decimal places.

Solution source

Let $t = \sinh^2 y$, then 

\begin{align*}
&& \sinh(2x) &= \cosh y  \tag{1}\\
&& \sinh(2y) &= 2 \cosh x \tag{2} \\ 
\\
 && \cosh(2x) &= 2 \cosh^2 x -1 \\
(2): &&&= \frac12 \sinh^2(2y) -1 \\
&& 1 &= \left (\frac12 \sinh^2(2y) -1 \right)^2 - \cosh^2 y \\
&&&= \frac14 \sinh^4(2y)-\sinh^2(2y)+1-\cosh^2 y \\
\Rightarrow && 0 &= \frac14 (\cosh^2 (2y)-1)^2- (\cosh^2 (2y)-1) - \cosh^2 y \\
&&&= \frac14 \left ( \left (1+2\sinh^2 y \right)^2-1 \right)^2 -\left ( \left (1+2\sinh^2 y \right)^2 -1\right) - (1 + \sinh^2 y ) \\
&&&= \frac14 \left ( 1 + 4t+4t^2 -1\right)^2 - \left ( 1+4t+4t^2-1\right) - (1 + t) \\
&&&= \frac14 (4t + 4t^2)^2 - (4t+4t^2)-1-t \\
&&&= 4(t+t^2)^2 - 4t^2-5t-1 \\
&&&= 4t^4+8t^3+4t^2-4t^2-5t-1 \\
&&& = 4t^4+8t^3-5t-1 \\
&&&= (t+1)(4t^3+4t^2-4t-1)
\end{align*}

Since $\sinh^2 y$ is positive, we must be a root of the second cubic. 

Let $f(t) = 4t^3+4t^2-4t-1$, then $f(0) = -1$ and $f'(t) = 12t^2+8t-4 = 4(t+1)(3t-1)$, so we have turning points at $-1$ and $\frac13$. Since $f(-1) = 3 > 0$ and $f(0) < 0$ we must have exactly one root larger than zero. Therefore there is a unique root.

$f(0.7) = -0.468 < 0$
$f(0.8) = 0.408 > 0$

since $f$ is continuous and changes sign, the root must fall in the interval $(0.7, 0.8)$.

Let $t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)}$, and $t_0 = 0.75$, then

\begin{align*}
t_0 &= 0.75 \\
t_1 &= 0.7571428571 \\
t_2 &= 0.7570684728 \\
t_3 &= 0.7570684647
\end{align*}

So $t \approx 0.757068$, $\sinh y \approx 0.870097$, $y \approx 0.786474$, $x \approx 0.546965$