If x = frac{npi}{2} satisfies the equation si | vedclass

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(B) Given the equation $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 - \sin x$. Let $t = \sin \frac{x}{2} - \cos \frac{x}{2}$. Then $t^2 = \sin^2 \frac{x}{

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$n = 1, 2, 4, 5$

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(B) Given the equation $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 - \sin x$. Let $t = \sin \frac{x}{2} - \cos \frac{x}{2}$. Then $t^2 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 - \sin x$. Substituting this into the equation,we get $t = t^2$,which implies $t(t - 1) = 0$. Thus,$\sin \frac{x}{2} - \cos \frac{x}{2} = 0$ or $\sin \frac{x}{2} - \cos \frac{x}{2} = 1$. Case $1$: $\sin \frac{x}{2} = \cos \frac{x}{2} \implies 	an \frac{x}{2} = 1 \implies \frac{x}{2} = k\pi + \frac{\pi}{4} \implies x = 2k\pi + \frac{\pi}{2}$. Given $x = \frac{n\pi}{2}$,we have $\frac{n\pi}{2} = 2k\pi + \frac{\pi}{2} \implies n = 4k + 1$. Case $2$: $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 \implies \sqrt{2} \sin(\frac{x}{2} - \frac{\pi}{4}) = 1 \implies \sin(\frac{x}{2} - \frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. This gives $\frac{x}{2} - \frac{\pi}{4} = 2m\pi + \frac{\pi}{4}$ or $\frac{x}{2} - \frac{\pi}{4} = 2m\pi + \frac{3\pi}{4}$. So,$\frac{x}{2} = 2m\pi + \frac{\pi}{2}$ or $\frac{x}{2} = 2m\pi + \pi$. Thus,$x = 4m\pi + \pi$ or $x = 4m\pi + 2\pi$. Given $x = \frac{n\pi}{2}$,we have $n = 8m + 2$ or $n = 8m + 4$. Checking the inequality $\left| \frac{x}{2} - \frac{\pi}{2} ight| \le \frac{3\pi}{4}$,we find that for $n \in \{1, 2, 4, 5\}$,the conditions are satisfied.

If $x = \frac{n\pi}{2}$ satisfies the equation $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 - \sin x$ and the inequality $\left| \frac{x}{2} - \frac{\pi}{2} \right| \le \frac{3\pi}{4}$,then:

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