If 4 sin 2 x – 8 sin x + 3 ≤ 0, 0 ≤ x ≤ 2π, then the solution set for x is
Mathematics · JEE Advanced · NTA Exams — Trigonometry
If 4 sin2x – 8 sin x + 3 ≤ 0, 0 ≤ x ≤ 2π, then the solution set for x is
- \(\left[0, \frac{\pi}{6}\right]\)
- \(\left[0, \frac{5 \pi}{6}\right]\)
- \(\left[\frac{5 \pi}{6}, 2 \pi\right]\)
- \(\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\)
Answer
(D) [ 6 , 5 6 ]
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