Calculus is not needed

Maximum value of the expression can be obtained when the value of the denominator is minimum. Applying AM-GM inequality, $256\sin^2 x+225\cos^2 x \geq480\sin x\cos x\\\large\frac{7440\sin x\cos x}{256\sin^2 x+225\cos^2 x}\leq\frac{7440}{480}\\\Huge\color{#3D99F6}{\leq\boxed{15.5}}$

Note first that

$(16\sin(x) - 15\cos(x))^{2} = 256\sin^{2}(x) + 225\cos^{2}(x) - 480\sin(x)\cos(x)$

$\Longrightarrow 7440\sin(x)\cos(x) = \dfrac{31}{2}*480\sin(x)\cos(x) =$

$\dfrac{31}{2}\left((256\sin^{2}(x) + 225\cos^{2}(x)) - (16\sin(x) - 15\cos(x))^{2}\right).$

Thus we can write the given expression as

$\dfrac{31}{2} - \dfrac{(16\sin(x) - 15\cos(x))^{2}}{256\sin^{2}(x) + 225\cos^{2}(x)} = \dfrac{31}{2} - A(x).$

Now since $A(x) \ge 0$ our original expression will achieve a maximum when $A(x) = 0,$ which is the case when

$16\sin(x) = 15\cos(x) \Longrightarrow \tan(x) = \frac{15}{16} \Longrightarrow x = \tan^{-1}(\frac{15}{16}),$

at which point the original expression achieves a maximum value of $\dfrac{31}{2} = \boxed{15.5}.$