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(D) Given $\sin 3x = \cos 2x$.We can write this as $\sin 3x = \sin \left( \frac{\pi}{2} - 2x \right)$.The general solution for $\sin A = \sin B$ is $A

Vedclass · Accepted Answer

$1$

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(D) Given $\sin 3x = \cos 2x$.We can write this as $\sin 3x = \sin \left( \frac{\pi}{2} - 2x \right)$.The general solution for $\sin A = \sin B$ is $A = n\pi + (-1)^n B$,where $n \in \mathbb{Z}$.Applying this,$3x = n\pi + (-1)^n \left( \frac{\pi}{2} - 2x \right)$.Case $1$: If $n$ is even,let $n = 2k$. Then $3x = 2k\pi + \frac{\pi}{2} - 2x$ $\Rightarrow 5x = 2k\pi + \frac{\pi}{2}$ $\Rightarrow x = \frac{2k\pi}{5} + \frac{\pi}{10} = \frac{(4k+1)\pi}{10}$.For $k=1$,$x = \frac{5\pi}{10} = \frac{\pi}{2}$ (not in interval).For $k=2$,$x = \frac{9\pi}{10}$ (in interval).Case $2$: If $n$ is odd,let $n = 2k+1$. Then $3x = (2k+1)\pi - \left( \frac{\pi}{2} - 2x \right) = 2k\pi + \pi - \frac{\pi}{2} + 2x = 2k\pi + \frac{\pi}{2} + 2x$ $\Rightarrow x = 2k\pi + \frac{\pi}{2}$.For $k=0$,$x = \frac{\pi}{2}$ (not in interval).For $k=1$,$x = 2\pi + \frac{\pi}{2} = \frac{5\pi}{2}$ (not in interval).Thus,the only solution in the interval $\left( \frac{\pi}{2}, \pi \right)$ is $x = \frac{9\pi}{10}$.Therefore,the number of solutions is $1$.

The number of solutions of $\sin 3x = \cos 2x$ in the interval $\left( \frac{\pi}{2}, \pi \right)$ is

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