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(A) Given equation: $\sin 2x + \cos x = 0$Using the identity $\sin 2x = 2 \sin x \cos x$,we get:$2 \sin x \cos x + \cos x = 0$Factoring out $\cos x$:$

Vedclass · Accepted Answer

$x = (2n+1)\frac{\pi}{2}$ or $x = n\pi + (-1)^n \frac{7\pi}{6}, n \in \mathbb{Z}$

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(A) Given equation: $\sin 2x + \cos x = 0$Using the identity $\sin 2x = 2 \sin x \cos x$,we get:$2 \sin x \cos x + \cos x = 0$Factoring out $\cos x$:$\cos x (2 \sin x + 1) = 0$This gives two cases:Case $1$: $\cos x = 0$The general solution for $\cos x = 0$ is $x = (2n + 1) \frac{\pi}{2}$,where $n \in \mathbb{Z}$.Case $2$: $2 \sin x + 1 = 0$$\sin x = -\frac{1}{2}$Since $\sin \frac{\pi}{6} = \frac{1}{2}$,we have $\sin x = -\sin \frac{\pi}{6} = \sin(\pi + \frac{\pi}{6}) = \sin \frac{7\pi}{6}$.The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$.Thus,$x = n\pi + (-1)^n \frac{7\pi}{6}$,where $n \in \mathbb{Z}$.Therefore,the general solution is $x = (2n + 1) \frac{\pi}{2}$ or $x = n\pi + (-1)^n \frac{7\pi}{6}$,where $n \in \mathbb{Z}$.

Find the general solution of the equation $\sin 2x + \cos x = 0$.

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