The number of solutions of the equation sin 2 | vedclass

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Question

(A) Given equation: $\sin 2x - 2 \cos x + 4 \sin x = 4$Using $\sin 2x = 2 \sin x \cos x$,we get:$2 \sin x \cos x - 2 \cos x + 4 \sin x - 4 = 0$Factor

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$3$

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(A) Given equation: $\sin 2x - 2 \cos x + 4 \sin x = 4$Using $\sin 2x = 2 \sin x \cos x$,we get:$2 \sin x \cos x - 2 \cos x + 4 \sin x - 4 = 0$Factor by grouping:$2 \cos x (\sin x - 1) + 4 (\sin x - 1) = 0$$(2 \cos x + 4)(\sin x - 1) = 0$$2(\cos x + 2)(\sin x - 1) = 0$Since $\cos x + 2 
eq 0$ for any real $x$ (as $-1 \le \cos x \le 1$),we must have:$\sin x = 1$In the interval $[0, 5\pi]$,the values of $x$ for which $\sin x = 1$ are:$x = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}$Thus,there are $3$ solutions.

The number of solutions of the equation $\sin 2x - 2 \cos x + 4 \sin x = 4$ in the interval $[0, 5\pi]$ is

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