The general value of theta satisfying the equ | vedclass

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(D) Given equation: $2\sin^2 	heta - 3\sin 	heta - 2 = 0$Factorizing the quadratic equation: $(2\sin 	heta + 1)(\sin 	heta - 2) = 0$Since $\sin

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$n\pi + (-1)^n \frac{7\pi}{6}$

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(D) Given equation: $2\sin^2 	heta - 3\sin 	heta - 2 = 0$Factorizing the quadratic equation: $(2\sin 	heta + 1)(\sin 	heta - 2) = 0$Since $\sin 	heta$ cannot be $2$,we have $2\sin 	heta + 1 = 0$,which implies $\sin 	heta = -\frac{1}{2}$.We know that $\sin 	heta = -\frac{1}{2} = \sin(-\frac{\pi}{6})$.The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$.Thus,$	heta = n\pi + (-1)^n (-\frac{\pi}{6})$.Using the property $(-1)^n (-1) = (-1)^{n+1}$,we can write $	heta = n\pi + (-1)^{n+1} \frac{\pi}{6}$.Alternatively,since $-\frac{\pi}{6}$ is equivalent to $\frac{11\pi}{6}$ or $\frac{7\pi}{6}$ in different contexts,checking the options,the form $n\pi + (-1)^n \frac{7\pi}{6}$ is mathematically consistent with the general solution set.

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ has two elements
$(II)$ $\{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$	$(Q)$ has three elements
$(III)$ $\{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos (2 x)=\sqrt{3}\}$	$(R)$ has four elements
$(IV)$ $\{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$	$(S)$ has five elements
	$(T)$ has six elements

The general value of $\theta$ satisfying the equation $2\sin^2 \theta - 3\sin \theta - 2 = 0$ is

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