Find the maximum value of $\sin (A) + \sin( 2A )+ \sin (3A)$

$$f(x)=\sin x+ \sin 2x+\sin 3x \implies f'(x)=12\cos^3x+4\cos^2 x-8\cos x-2]$$ $f'(x)=0$ has three real roots in $[0,2\pi]$ these are $\cos x=-0.8768, -0.2419, 0.7855$$ \implies x=2.6399...,1.8151...,0.6672...$, respectively. the respective values of $f(x)$ are $0.6354..., -0.2423..., 2.4996...$ Finally $f_{max}=f(0.6672...)=2.4996..$

According to your graph, the maximum seems to appear "close" to $A=\frac \pi 5$. Develop the expression as a Taylor series to get for the function $$\sqrt{\frac{25}{8}+\frac{11 \sqrt{5}}{8}}+\frac{1}{2} \left(A-\frac{\pi }{5}\right)-\frac{1}{4} \sqrt{5 \left(85+22 \sqrt{5}\right)} \left(A-\frac{\pi }{5}\right)^2+O\left(\left(A-\frac{\pi }{5}\right)^3\right)$$ Ignoring the higher order terms, the derivative cancels for $$A=\frac \pi 5+\frac{1}{\sqrt{5 \left(85+22 \sqrt{5}\right)}}\approx 0.666924$$ and for this value the truncated series gives a maximum of $$\frac{\sqrt{5}+5 \sqrt{6670+2970 \sqrt{5}}}{20 \sqrt{85+22 \sqrt{5}}}\approx 2.49955$$

The exact solution of the problem is $(0.667291,2.49961)$. Not too bad for a solution obtained without solving any equation beside a linear one.