The number of solutions in [0, frac{pi}{2}) o | vedclass

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(D) Given equation: $\cos 3x \cdot 	an 5x = \sin 7x$$\cos 3x \cdot \frac{\sin 5x}{\cos 5x} = \sin 7x$$\cos 3x \cdot \sin 5x = \sin 7x \cdot \cos 5x$U

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$6$

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(D) Given equation: $\cos 3x \cdot 	an 5x = \sin 7x$$\cos 3x \cdot \frac{\sin 5x}{\cos 5x} = \sin 7x$$\cos 3x \cdot \sin 5x = \sin 7x \cdot \cos 5x$Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$\frac{1}{2} [\sin(5x+3x) + \sin(5x-3x)] = \frac{1}{2} [\sin(7x+5x) + \sin(7x-5x)]$$\sin 8x + \sin 2x = \sin 12x + \sin 2x$$\sin 12x = \sin 8x$$\sin 12x - \sin 8x = 0$$2 \cos(10x) \sin(2x) = 0$Case $1$: $\sin 2x = 0 \implies 2x = n\pi \implies x = \frac{n\pi}{2}$. In $[0, \frac{\pi}{2})$, $x = 0$.Case $2$: $\cos 10x = 0 \implies 10x = (2n+1)\frac{\pi}{2} \implies x = \frac{(2n+1)\pi}{20}$.For $x \in [0, \frac{\pi}{2})$, $n = 0, 1, 2, 3, 4 \implies x = \frac{\pi}{20}, \frac{3\pi}{20}, \frac{5\pi}{20}, \frac{7\pi}{20}, \frac{9\pi}{20}$.Check domain: $	an 5x$ is undefined at $5x = (2k+1)\frac{\pi}{2} \implies x = \frac{(2k+1)\pi}{10}$.For $k=0, x=\frac{\pi}{10} = \frac{2\pi}{20}$ (not in set); $k=1, x=\frac{3\pi}{10} = \frac{6\pi}{20}$ (not in set); $k=2, x=\frac{5\pi}{10} = \frac{10\pi}{20}$ (not in set).Valid solutions: $x = 0, \frac{\pi}{20}, \frac{3\pi}{20}, \frac{5\pi}{20}, \frac{7\pi}{20}, \frac{9\pi}{20}$. Total $6$ solutions.

The number of solutions in $[0, \frac{\pi}{2})$ of the equation $\cos 3x \cdot \tan 5x = \sin 7x$ is

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