Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4 \sin 2x}{e^{\sqrt{x-1}}}, \quad 0 \leq x \leq \pi$ The curve has a maximum turning point at $P$ and a minimum turning point at $Q$ as shown in Figure 5 - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 2

To show that the $x$ coordinates correspond to solutions of the equation, we first differentiate the function:

1. Differentiate the Function:
   Using the quotient rule, we have:
   $f'(x) = \frac{(e^{\sqrt{x-1}})(8 \cos 2x) - (4 \sin 2x)(\frac{1}{2\sqrt{x-1}} e^{\sqrt{x-1}})}{(e^{\sqrt{x-1}})^2}$

2. Set the Derivative to Zero:
   Setting the derivative equal to zero and simplifying gives us the conditions for maximum and minimum points.
   This results in the use of the trigonometric identity.

3. Solve for $x$:
   We proceed by solving the equation
   $\tan 2x = \sqrt{2}$
   This gives us critical points which are angles where the tangent equals rac{\sqrt{2}}{1}. The basic solution can be noted as: $2x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}$
   Dividing through by 2 leads to: $x = \frac{\pi}{8} + \frac{n\pi}{2}$.
   From the domain $0 \leq x \leq \pi$, we identify valid $x$ coordinates for points $P$ and $Q$.