If |sin x-cos ^2 x| geq|3-3 sin x+sin ^2 x|+4 | vedclass

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$(4 n+1) \frac{\pi}{2}, n \in Z$

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(A) Given the inequality: $|\sin x-\cos ^2 x| \geq|3-3 \sin x+\sin ^2 x|+4|\sin x-1|$Substitute $\cos ^2 x = 1-\sin ^2 x$:$|\sin x-(1-\sin ^2 x)| \geq|\sin ^2 x-3 \sin x+3|+4|\sin x-1|$$|\sin ^2 x+\sin x-1| \geq|\sin ^2 x-3 \sin x+3|+|4 \sin x-4|$Note that $(\sin ^2 x-3 \sin x+3) + (4 \sin x-4) = \sin ^2 x+\sin x-1$.Let $a = \sin ^2 x-3 \sin x+3$ and $b = 4 \sin x-4$. The inequality is $|a+b| \geq |a|+|b|$.This holds if and only if $ab \geq 0$.Since the discriminant of $a = \sin ^2 x-3 \sin x+3$ is $D = (-3)^2 - 4(1)(3) = 9-12 = -3  0$ for all $x \in R$.Thus,$b \geq 0$ $\Rightarrow 4(\sin x-1) \geq 0$ $\Rightarrow \sin x \geq 1$.Since $\sin x \leq 1$,we must have $\sin x = 1$.Therefore,$x = 2n\pi + \frac{\pi}{2} = (4n+1)\frac{\pi}{2}, n \in Z$.

If $|\sin x-\cos ^2 x| \geq|3-3 \sin x+\sin ^2 x|+4|\sin x-1|$,then $x=$

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