Question:

<p><strong>Step 1: Set up and solve the quadratic equation.</strong></p><p>Let $y = \sin x$. The equation becomes:</p><p>$$3y^2 - 7y + 2 = 0$$</p><p><strong>Step 2: Factor or use the quadratic formula.</strong></p><p>Using the quadratic formula:</p><p>$$y = \frac{7 \pm \sqrt{49 - 24}}{6} = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6}$$</p><p>This gives: $y = \frac{12}{6} = 2$ or $y = \frac{2}{6} = \frac{1}{3}$</p><p><strong>Step 3: Check validity of solutions.</strong></p><p>Since $\sin x$ must satisfy $-1 \leq \sin x \leq 1$:</p><p>• $\sin x = 2$ is impossible (outside the range)</p><p>• $\sin x = \frac{1}{3}$ is valid ✓</p><p><strong>Step 4: Count solutions for $\sin x = \frac{1}{3}$ in $[0, 5\pi]$.</strong></p><p>In one period $[0, 2\pi]$, the equation $\sin x = \frac{1}{3}$ has exactly 2 solutions:</p><p>• One in $[0, \pi]$ at some angle $\alpha$ where $\sin \alpha = \frac{1}{3}$</p><p>• One in $[\pi, 2\pi]$ at $\pi - \alpha$ (since $\sin(\pi - \alpha) = \sin \alpha$)</p><p><strong>Step 5: Determine the interval structure.</strong></p><p>The interval $[0, 5\pi]$ contains:</p><p>• 2 complete periods: $[0, 2\pi]$ and $[2\pi, 4\pi]$ → 2 × 2 = 4 solutions</p><p>• Partial period: $[4\pi, 5\pi]$ has length $\pi$ (exactly half a period)</p><p>In $[4\pi, 5\pi]$, which corresponds to $[0, \pi]$ modulo $2\pi$, there is exactly 1 solution where $\sin x = \frac{1}{3}$</p><p><strong>Step 6: Sum all solutions.</strong></p><p>Total number of solutions = 4 + 1 + 1 = 6</p><p>Alternatively: $\frac{5\pi}{2\pi} = 2.5$ periods, and each period contributes 2 solutions: $2 × 2 + 1 = 5$ solutions from 2 full periods, plus 1 from the remaining half period = 6 solutions.</p><p>∴ Answer: C</p>