$$\max _\limits{0 \leq x \leq \pi}\left\{x-2 \sin x \cos x+\frac{1}{3} \sin 3 x\right\}=$$

<p>Given the function:</p> <p>$f(x) = x - 2\sin{x}\cos{x} + \frac{1}{3}\sin{3x}$</p> <p>We want to find the maximum value of this expression for $0 \leq x \leq \pi$.</p> <p>Step 1: Rewrite the expression <br/><br/>Notice that we can rewrite the expression as: <br/><br/>$f(x) = x - \sin{2x} + \frac{1}{3}\sin{3x}$</p> <p>Step 2: Find the first derivative <br/><br/>Now, let&#39;s find the derivative of this expression with respect to $x$: <br/><br/>$f&#39;(x) = 1 - 2\cos{2x} + \cos{3x}$</p> <p>Step 3: Find the critical points <br/><br/>To find the critical points, we set $f&#39;(x) = 0$: <br/><br/>$1 - 2\cos{2x} + \cos{3x} = 0$</p> <p>Step 4: Solve for x <br/><br/>This equation can be rewritten as: <br/><br/>$(2\cos{x} + \sqrt{3})(2\cos{x} - \sqrt{3})(\cos{x} - 1) = 0$</p> <p>We get three possible solutions for $x$: <br/><br/>$\cos{x} = \frac{-\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 1$</p> <p>Which gives us: $x = \frac{5\pi}{6}, \frac{\pi}{6}, 0$</p> <p>Step 5: Find the second derivative <br/><br/>Now, let&#39;s find the second derivative of the expression: <br/><br/>$f&#39;&#39;(x) = 4\sin{2x} - 3\sin{3x}$</p> <p>Step 6: Analyze the critical points Evaluate the second derivative at the critical points:</p> <ol> <li>$f&#39;&#39;\left(\frac{5\pi}{6}\right) = -2\sqrt{3} - \sqrt{3} &lt; 0$, indicating a local maximum.</li> <li>$f&#39;&#39;\left(\frac{\pi}{6}\right) = 2\sqrt{3} - \sqrt{3} &gt; 0$, indicating a local minimum.</li> <li>$f&#39;&#39;(0) = 0$, inconclusive.</li> </ol> <p>Step 7: Evaluate the function at the local maximum point and the boundary points Since we are looking for the maximum value of $f(x)$, we can now evaluate the function at the local maximum point and the boundary points:</p> <ol> <li>$$f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6}$$</li> <li>$f(0) = 0$</li> <li>$f(\pi) = \pi$</li> </ol> <p>Step 8: Compare the values The maximum value of $f(x)$ is given by $f\left(\frac{5\pi}{6}\right) = \frac{5\pi + 2 + 3\sqrt{3}}{6}$.</p>