If the possible solutions of the equation 2 c | vedclass

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(B) Given equation: $2 \cos ^2 x + 3 \sin x - 3 = 0$Using the identity $\cos ^2 x = 1 - \sin ^2 x$,we get:$2(1 - \sin ^2 x) + 3 \sin x - 3 = 0$$2 - 2

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$\frac{\pi}{3}$

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(B) Given equation: $2 \cos ^2 x + 3 \sin x - 3 = 0$Using the identity $\cos ^2 x = 1 - \sin ^2 x$,we get:$2(1 - \sin ^2 x) + 3 \sin x - 3 = 0$$2 - 2 \sin ^2 x + 3 \sin x - 3 = 0$$-2 \sin ^2 x + 3 \sin x - 1 = 0$$2 \sin ^2 x - 3 \sin x + 1 = 0$Factoring the quadratic equation:$(2 \sin x - 1)(\sin x - 1) = 0$This gives $\sin x = \frac{1}{2}$ or $\sin x = 1$.For $\sin x = \frac{1}{2}$,$x = \frac{\pi}{6}$ (or $30^\circ$).For $\sin x = 1$,$x = \frac{\pi}{2}$ (or $90^\circ$).The two unequal angles are $\frac{\pi}{6}$ and $\frac{\pi}{2}$.The sum of angles in a triangle is $\pi$.Third angle $= \pi - (\frac{\pi}{6} + \frac{\pi}{2}) = \pi - (\frac{\pi + 3\pi}{6}) = \pi - \frac{4\pi}{6} = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$.

If the possible solutions of the equation $2 \cos ^2 x + 3 \sin x - 3 = 0$ constitute two unequal angles of a triangle,then the third angle of that triangle is

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