1. **Problem Statement:** Apply Rolle’s theorem to the function $f(x) = \sin x \sqrt{\cos 2x}$ on the interval $[0, \frac{\pi}{4}]$ and find $x$ such that $0 < x < \frac{\pi}{4}$ where $f'(x) = 0$. 2. **Rolle’s Theorem:** If a function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$. 3. **Check conditions:** - $f(x)$ is continuous on $[0, \frac{\pi}{4}]$ because $\sin x$ and $\sqrt{\cos 2x}$ are continuous there. - $f(x)$ is differentiable on $(0, \frac{\pi}{4})$. - Calculate $f(0)$ and $f(\frac{\pi}{4})$: $$f(0) = \sin 0 \cdot \sqrt{\cos 0} = 0 \cdot 1 = 0$$ $$f\left(\frac{\pi}{4}\right) = \sin \frac{\pi}{4} \cdot \sqrt{\cos \frac{\pi}{2}} = \frac{\sqrt{2}}{2} \cdot \sqrt{0} = 0$$ So, $f(0) = f(\frac{\pi}{4}) = 0$. 4. **Find $f'(x)$:** Let $f(x) = \sin x \cdot (\cos 2x)^{1/2}$. Using product rule: $$f'(x) = \cos x \cdot \sqrt{\cos 2x} + \sin x \cdot \frac{1}{2} (\cos 2x)^{-1/2} \cdot (-2 \sin 2x)$$ Simplify: $$f'(x) = \cos x \sqrt{\cos 2x} - \sin x \frac{\sin 2x}{\sqrt{\cos 2x}}$$ 5. **Set $f'(x) = 0$ and solve for $x$ in $(0, \frac{\pi}{4})$:** $$\cos x \sqrt{\cos 2x} = \sin x \frac{\sin 2x}{\sqrt{\cos 2x}}$$ Multiply both sides by $\sqrt{\cos 2x}$: $$\cos x \cos 2x = \sin x \sin 2x$$ Rewrite as: $$\cos x \cos 2x - \sin x \sin 2x = 0$$ Use cosine addition formula: $$\cos(x + 2x) = \cos 3x = 0$$ 6. **Solve $\cos 3x = 0$ for $x$ in $(0, \frac{\pi}{4})$:** $$3x = \frac{\pi}{2} \implies x = \frac{\pi}{6}$$ This is the only solution in $(0, \frac{\pi}{4})$. **Final answer:** $$x = \frac{\pi}{6}$$