Question 46, Find the maximum value of the function $\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\sqrt{3} \cos \mathrm{x}-2 \sin 3 \mathrm{x}$.
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\begin{array}{l}
f(x)=\sin x+\sqrt{3} \cos x-2 \sin 3 x=2 \sin \left(x+\frac{\pi}{3}\right)+2 \sin 3\left(x+\frac{\pi}{3}\right) \\
\text { Let } \alpha=x+\frac{\pi}{3} \text {, then } \\
f(x)=2(\sin \alpha+\sin 3 \alpha)=2\left(\sin \alpha+3 \sin \alpha-4 \sin ^{3} \alpha\right)=8 \sin \alpha \cos ^{2} \alpha \\
=4 \sqrt{2} \sqrt{2 \sin ^{2} \alpha \cdot \cos ^{2} \alpha \cdot \cos ^{2} \alpha} \leq 4 \sqrt{2} \sqrt{\left(\frac{2 \sin ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha}{3}\right)^{3}}=\frac{16 \sqrt{3}}{9} \\
\text { It is clear that equality can be achieved. }
\end{array}
$$