The number of roots of the equation cos^2 x + | vedclass

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(B) Given equation: $\cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$Substitute $\cos^2 x = 1 - \sin^2 x$:$1 - \sin^2 x + \frac{

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$4$

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(B) Given equation: $\cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$Substitute $\cos^2 x = 1 - \sin^2 x$:$1 - \sin^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$$-\sin^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} = 0$Multiply by $-4$:$4 \sin^2 x - 2(\sqrt{3} + 1) \sin x + \sqrt{3} = 0$$4 \sin^2 x - 2\sqrt{3} \sin x - 2 \sin x + \sqrt{3} = 0$$2 \sin x (2 \sin x - \sqrt{3}) - 1(2 \sin x - \sqrt{3}) = 0$$(2 \sin x - 1)(2 \sin x - \sqrt{3}) = 0$So,$\sin x = \frac{1}{2}$ or $\sin x = \frac{\sqrt{3}}{2}$.For $\sin x = \frac{1}{2}$ in $[-\pi, \pi]$,$x = \frac{\pi}{6}, \frac{5\pi}{6}$.For $\sin x = \frac{\sqrt{3}}{2}$ in $[-\pi, \pi]$,$x = \frac{\pi}{3}, \frac{2\pi}{3}$.Total number of roots is $4$.

The number of roots of the equation $\cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$ that lie in the interval $[-\pi, \pi]$ is:

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