Question 45: Given $x \in R$, find the maximum value of $\frac{\sin x(2-\cos x)}{5-4 \cos x}$.

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Question 45, Solution Method One: Let $M=\frac{\sin x(2-\cos x)}{5-4 \cos x}$, and let $a=5-4 \cos x$, then $a \in[1, 9]$,
$$
\begin{array}{l}
M^{2}=\frac{\left(1-\cos ^{2} x\right)(2-\cos x)^{2}}{(5-4 \cos x)^{2}}=\frac{\left(1-\cos ^{2} x\right)(2-\cos x)^{2}}{(5-4 \cos x)^{2}}=\frac{\left[1-\left(\frac{5-a}{4}\right)^{2}\right]\left(2 \frac{5-a}{4}\right)^{2}}{a^{2}}=\frac{1}{16^{2}} \cdot \frac{(9-a)(a-1)(3+a)^{2}}{a^{2}} \\
\leq \frac{1}{16} \cdot \frac{\left.\left[\frac{(9-a)+(3+a)}{2}\right]^{2} \times \frac{1}{3} \frac{3(a-1)+(3+a)}{2}\right]^{2}}{a^{2}}=\frac{1}{16^{2}} \cdot \frac{6^{2} \times \frac{1}{3}(2 a)^{2}}{a^{2}}=\frac{3}{16}
\end{array}
$$

Therefore, $M \leq \frac{\sqrt{3}}{4}$, with equality when $a=3 \Leftrightarrow \cos x=\frac{1}{2}$.
Given $x \in \mathbb{R}$, find the maximum value of $\frac{\sin x(2-\cos x)}{5-4 \cos x}$.
Solution Method Two: Let $M=\frac{\sin x(2-\cos x)}{5-4 \cos x}$, and let $a=5-4 \cos x$, then $a \in[1, 9]$, and
$$
\begin{array}{l}
\text { Solution Method Two: Let } M=\frac{\sin x(2-\cos x)}{5-4 \cos x}, \text { and let } a=5-4 \cos x, \text { then } a \in[1, 9], \text { and } \\
M^{2}=\frac{\left(1-\cos ^{2} x\right)(2-\cos x)^{2}}{(5-4 \cos x)^{2}}=\frac{\left(1-\cos ^{2} x\right)(2-\cos x)^{2}}{(5-4 \cos x)^{2}}=\frac{\left[1-\left(\frac{5-a}{4}\right)^{2}\right]\left(2-\frac{5-2}{4}\right)^{2}}{a^{2}}=\frac{1}{16^{2}} \cdot \frac{(9-a)(a-1)(3+a)^{2}}{a^{2}} \\
=\frac{1}{16^{2}} \cdot\left(10-\left(a+\frac{9}{a}\right)\right)\left(a+\frac{9}{a}+6\right) \leq \frac{3}{16}
\end{array}
$$

Therefore, $M \leq \frac{\sqrt{3}}{4}$, with equality when $a=3 \Leftrightarrow \cos x=\frac{1}{2}$.
Solution Method Three: Let $M=\frac{\sin x(2-\cos x)}{5-4 \cos x}$, then
$$
M^{2}=\frac{\sin ^{2} x(2-\cos x)^{2}}{\left(\sin ^{2} x+(2-\cos x)^{2}\right)^{2}}=\frac{\left(\frac{\sin x}{2-\cos x}\right)^{2}}{\left(\left(\frac{\sin x}{2-\cos x}\right)^{2}+1\right)^{2}}=\frac{1}{\left(\frac{\sin x}{2-\cos x}\right)^{2}+2+\frac{1}{\left(\frac{\sin x}{2-\cos x}\right)^{2}}}
$$

Let $t=\frac{\sin x}{2-\cos x}$, then $2 t=t \cdot \cos x+\sin x \leq \sqrt{t^{2}+1} \Rightarrow t^{2} \leq \frac{1}{3}$, hence
$$
M^{2}=\frac{1}{t^{2}+2+\frac{1}{t^{2}}} \leq \frac{1}{\frac{1}{3}+2+3}=\frac{3}{16}
$$

Therefore, $M \leq \frac{\sqrt{3}}{4}$, with equality when $a=3 \Leftrightarrow \cos x=\frac{1}{2}$.