Question:medium

Set of values of \(x\) lying in \([0,2\pi]\) satisfying the inequality \( |\sin x|>2\sin^2 x \) contains:

Show Hint

Split modulus inequalities into cases — positive and negative parts.
Updated On: Apr 17, 2026
  • \( (0,\frac{\pi}{6}) \cup (\pi,\frac{7\pi}{6}) \)
  • \( (0,\frac{7\pi}{6}) \)
  • \( \frac{\pi}{6} \)
  • None of these
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The inequality involves an absolute value and a square. Since \(\sin^{2} x = |\sin x|^{2}\), we can rewrite the inequality in terms of \(|\sin x|\).
Step 2: Key Formula or Approach:
Let \(t = |\sin x|\). Then the inequality is \(t>2t^{2}\).
Step 3: Detailed Explanation:
1. Solve for \(t\):
\[ t - 2t^{2}>0 \]
\[ t(1 - 2t)>0 \]
Since \(t = |\sin x|\) must be non-negative, and for the strict inequality to hold, \(t\) must be strictly positive (\(\sin x \neq 0\)).
\[ 1 - 2t>0 \implies t<1/2 \]
2. Thus, the condition is \(0<|\sin x|<1/2\).
3. This implies:
- For \(x \in [0, \pi]\): \(0<\sin x<1/2 \implies x \in (0, \pi/6) \cup (5\pi/6, \pi)\).
- For \(x \in [\pi, 2\pi]\): \(0<-\sin x<1/2 \implies -1/2<\sin x<0 \implies x \in (\pi, 7\pi/6) \cup (11\pi/6, 2\pi)\).
4. Checking the options: Option (A) provides the intervals \((0, \pi/6)\) and \((\pi, 7\pi/6)\), which are part of the solution set.
Step 4: Final Answer:
The set of values contains \((0, \pi/6) \cup (\pi, 7\pi/6)\).
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