The number of solutions of the equation sin(9 | vedclass

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(B) Given equation: $\sin(9x) + \sin(3x) = 0$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$2 \sin(6x

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$13$

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(B) Given equation: $\sin(9x) + \sin(3x) = 0$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$2 \sin(6x) \cos(3x) = 0$This implies $\sin(6x) = 0$ or $\cos(3x) = 0$.Case $1$: $\sin(6x) = 0$ $\Rightarrow 6x = n\pi$ $\Rightarrow x = \frac{n\pi}{6}$ for $n \in \mathbb{Z}$.For $x \in [0, 2\pi]$,$n$ can be $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$.Values: $x \in \{0, \frac{\pi}{6}, \frac{2\pi}{6}, \frac{3\pi}{6}, \frac{4\pi}{6}, \frac{5\pi}{6}, \frac{6\pi}{6}, \frac{7\pi}{6}, \frac{8\pi}{6}, \frac{9\pi}{6}, \frac{10\pi}{6}, \frac{11\pi}{6}, \frac{12\pi}{6}\}$ (Total $13$ values).Case $2$: $\cos(3x) = 0$ $\Rightarrow 3x = (2k+1)\frac{\pi}{2}$ $\Rightarrow x = (2k+1)\frac{\pi}{6}$ for $k \in \mathbb{Z}$.For $x \in [0, 2\pi]$,$k$ can be $0, 1, 2, 3, 4, 5$.Values: $x \in \{\frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{9\pi}{6}, \frac{11\pi}{6}\}$.All these values are already included in the set from Case $1$.Thus,the total number of distinct solutions is $13$.

The number of solutions of the equation $\sin(9x) + \sin(3x) = 0$ in the closed interval $[0, 2\pi]$ is

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