The number of solutions of the equation (4-sq | vedclass

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(D) Given equation: $(4-\sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}$Rationalizing the $RHS$: $-\frac{4(1-\sqrt{3})}{(1+\sqrt{3})(1-

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$5$

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(D) Given equation: $(4-\sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}$Rationalizing the $RHS$: $-\frac{4(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} = -\frac{4(1-\sqrt{3})}{1-3} = 2(1-\sqrt{3}) = 2 - 2\sqrt{3}$Substitute $\cos^2 x = 1 - \sin^2 x$:$(4-\sqrt{3}) \sin x - 2\sqrt{3}(1 - \sin^2 x) = 2 - 2\sqrt{3}$$2\sqrt{3} \sin^2 x + (4-\sqrt{3}) \sin x - 2\sqrt{3} = 2 - 2\sqrt{3}$$2\sqrt{3} \sin^2 x + (4-\sqrt{3}) \sin x - 2 = 0$$(2 \sin x - 1)(\sqrt{3} \sin x + 2) = 0$Since $\sin x = -\frac{2}{\sqrt{3}}$ is impossible,we have $\sin x = \frac{1}{2}$.In the interval $x \in [-2\pi, \frac{5\pi}{2}]$,the values of $x$ for which $\sin x = \frac{1}{2}$ are:$x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}$$x = -2\pi + \frac{5\pi}{6} = -\frac{7\pi}{6}$$x = \frac{\pi}{6}$$x = \frac{5\pi}{6}$$x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$Total number of solutions is $5$.

The number of solutions of the equation $(4-\sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}$ for $x \in [-2\pi, \frac{5\pi}{2}]$ is

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