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Question: Answered & Verified by Expert
If $\left|\sin x-\cos ^2 x\right| \geq\left|3-3 \sin x+\sin ^2 x\right|+4|\sin x-1|$, then $x=$
MathematicsTrigonometric EquationsTS EAMCETTS EAMCET 2020 (11 Sep Shift 1)
Options:
  • A $(4 n+1) \frac{\pi}{2}, n \in Z$
  • B $2 n \pi+\frac{\pi}{3}, n \in Z$
  • C $n \pi+\frac{\pi}{2}, n \in Z$
  • D $2 n \pi+\frac{\pi}{6}, n \in Z$
Solution:
2944 Upvotes Verified Answer
The correct answer is: $(4 n+1) \frac{\pi}{2}, n \in Z$
It is given that,
$\left|\sin x-\cos ^2 x\right| \geq\left|3-3 \sin x+\sin ^2 x\right|+4|\sin x-1|$
$\Rightarrow\left|\sin ^2 x+\sin x-1\right| \geq\left|\sin ^2 x-3 \sin x+3\right|$ $+|4 \sin x-4|$
$\Rightarrow\left|\left(\sin ^2 x-3 \sin x+3\right)+(4 \sin x-4)\right|$ $\geq\left|\sin ^2 x-3 \sin x+3\right|+|4 \sin x-4|$
$\because$ If $|a+b| \geq|a|+|b| \Rightarrow a b \geq 0$
$\therefore\left(\sin ^2 x-3 \sin x+3\right)(4 \sin x-4) \geq 0$
$\because \sin ^2 x-3 \sin x+3>0 \forall x \in \mathbf{R}$
$\therefore \sin x-1 \geq 0 \Rightarrow \sin x=1 \Rightarrow x=(4 n+1) \frac{\pi}{2}, n \in \mathbf{Z}$

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