Find 9A

Let $f(x)$ be the given function. Then

$\begin{aligned} f(x) & = \frac {\sin 2x}{\sqrt{19\sin^2 x + 10 \cos^2 x +6}} & \small \blue{\text{Since }\sin^2 \theta + \cos^2 \theta = 1} \\ & = \frac {\sin 2x}{\sqrt{9\sin^2 x + 16}} & \small \blue{\text{and }\sin^2 \theta = \frac {1-\cos 2\theta}2} \\ & = \frac {\sqrt 2 \sin 2x}{\sqrt{41-9\cos 2x}} \\ \frac {df(x)}{dx} & = \frac {2 \sqrt 2 \cos 2 x}{\sqrt{41-9\cos 2x}} - \frac {9\sqrt 2 \sin^2 2x}{(41-9\cos 2x)^\frac 32} \\ & = \frac {2 \sqrt 2 \cos 2x (41-9\cos 2x) - 9\sqrt 2 \sin^2 2x}{(41-9\cos 2x)^\frac 32} \end{aligned}$

Putting $\dfrac {df(x)}{dx} = 0$ ,

$\begin{aligned} 82 \cos 2x - 18 \cos^2 2x - 9 \sin^2 2x & = 0 & \small \blue{\text{Since }\sin^2 \theta + \cos^2 \theta = 1} \\ 9 \cos^2 2x - 82 \cos 2x + 9 & = 0 \\ (9\cos 2x - 1) (\cos 2x - 9) & = 0 \\ \implies \cos 2x & = \frac 19 & \small \blue{\text{Since }|\cos \theta| \le 1} \end{aligned}$

When $\cos 2x = \dfrac 19$ , $\implies \sin 2x = \pm \dfrac {4\sqrt 5}9$ . SInce $f(x)$ is periodical, $f(x)$ is maximum when $\sin 2x = \dfrac {\sqrt {80}}9$ and minimum when $\sin 2x = - \dfrac {4\sqrt 5}9$ . Therefore,

$\begin{aligned} A & = \max(f(x)) = \frac {\sqrt 2 \sin 2x}{\sqrt{41-9\cos 2x}}\bigg|_{\sin 2x = \frac {\sqrt{80}}9} = \frac {\sqrt{160}}{9\sqrt{40}} = \frac 29 \end{aligned}$

Therefore $9A = \boxed 2$ .