1. **State the problem:** We need to find the exact value of $a$ where the curve $y = \cos x \sqrt{\sin 2x}$ has a maximum point $M$ for $0 \leq x \leq \frac{\pi}{2}$. 2. **Write the function:** $$y = \cos x \sqrt{\sin 2x}$$ 3. **Find the derivative $y'$ to locate critical points:** Let $u = \cos x$ and $v = \sqrt{\sin 2x} = (\sin 2x)^{1/2}$. Using the product rule: $$y' = u'v + uv'$$ 4. **Calculate $u'$ and $v'$:** $$u' = -\sin x$$ For $v = (\sin 2x)^{1/2}$, use the chain rule: $$v' = \frac{1}{2}(\sin 2x)^{-1/2} \cdot \cos 2x \cdot 2 = \frac{\cos 2x}{\sqrt{\sin 2x}}$$ 5. **Substitute back into $y'$:** $$y' = (-\sin x) \sqrt{\sin 2x} + \cos x \cdot \frac{\cos 2x}{\sqrt{\sin 2x}} = \frac{-\sin x (\sin 2x) + \cos x \cos 2x}{\sqrt{\sin 2x}}$$ 6. **Set numerator equal to zero to find critical points:** $$-\sin x (\sin 2x) + \cos x \cos 2x = 0$$ 7. **Rewrite using $\sin 2x = 2 \sin x \cos x$:** $$-\sin x (2 \sin x \cos x) + \cos x \cos 2x = 0$$ $$-2 \sin^2 x \cos x + \cos x \cos 2x = 0$$ 8. **Factor out $\cos x$:** $$\cos x (-2 \sin^2 x + \cos 2x) = 0$$ 9. **Solve each factor:** - $\cos x = 0$ gives $x = \frac{\pi}{2}$ (boundary, check later) - $-2 \sin^2 x + \cos 2x = 0$ 10. **Use identity $\cos 2x = 1 - 2 \sin^2 x$:** $$-2 \sin^2 x + 1 - 2 \sin^2 x = 0$$ $$1 - 4 \sin^2 x = 0$$ 11. **Solve for $\sin^2 x$:** $$4 \sin^2 x = 1$$ $$\sin^2 x = \frac{1}{4}$$ $$\sin x = \frac{1}{2}$$ (since $0 \leq x \leq \frac{\pi}{2}$, sine is positive) 12. **Find $x$:** $$x = \frac{\pi}{6}$$ 13. **Check values at critical points and boundaries:** - At $x=0$, $y=\cos 0 \sqrt{\sin 0} = 1 \cdot 0 = 0$ - At $x=\frac{\pi}{6}$, $$y = \cos \frac{\pi}{6} \sqrt{\sin \frac{\pi}{3}} = \frac{\sqrt{3}}{2} \cdot \sqrt{\frac{\sqrt{3}}{2}}$$ - At $x=\frac{\pi}{2}$, $y=\cos \frac{\pi}{2} \sqrt{\sin \pi} = 0 \cdot 0 = 0$ 14. **Since $y$ is positive and nonzero at $x=\frac{\pi}{6}$ and zero at boundaries, the maximum is at** $$a = \frac{\pi}{6}$$