The general solution of the equation sqrt{3-5 | vedclass

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(C) Given equation: $\sqrt{3-5 \sin x+\sin ^2 x} = -\cos x$.For the square root to be defined and equal to $-\cos x$,we must have $-\cos x \ge 0$,whic

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$(2 n+1) \pi-\frac{\pi}{6}, n \in Z$

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(C) Given equation: $\sqrt{3-5 \sin x+\sin ^2 x} = -\cos x$.For the square root to be defined and equal to $-\cos x$,we must have $-\cos x \ge 0$,which implies $\cos x \le 0$.Squaring both sides: $3-5 \sin x+\sin ^2 x = \cos ^2 x$.Using $\cos ^2 x = 1 - \sin ^2 x$: $3-5 \sin x+\sin ^2 x = 1 - \sin ^2 x$.$2 \sin ^2 x - 5 \sin x + 2 = 0$.Factoring the quadratic: $(2 \sin x - 1)(\sin x - 2) = 0$.This gives $\sin x = \frac{1}{2}$ or $\sin x = 2$.Since $\sin x \in [-1, 1]$,we must have $\sin x = \frac{1}{2}$.For $\sin x = \frac{1}{2}$,$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$.Since $\cos x \le 0$,we must have $x = \frac{5\pi}{6}$ in the interval $[0, 2\pi]$.The general solution for $\sin x = \frac{1}{2}$ and $\cos x < 0$ is $x = 2n\pi + \frac{5\pi}{6}$,which is equivalent to $x = (2n+1)\pi - \frac{\pi}{6}$.

The general solution of the equation $\sqrt{3-5 \sin x+\sin ^2 x}+\cos x=0$ is

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