The solution set of $ x \in(-\pi, \pi) $ for the inequality $ \sin 2 x+1 \leq \cos x+2 \sin x $ is

Question

The solution set of $ x \in(-\pi, \pi) $ for the inequality $ \sin 2 x+1 \leq \cos x+2 \sin x $ is, $ x \in\left[0, \frac{\pi}{6}\right] $, $ x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right] \cup\{0\} $, $ x \in\left[-\frac{\pi}{6}, \frac{5 \pi}{6}\right] $, $ x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $

MARKS App · Accepted Answer

2 sin ⁡ x cos ⁡ x + 1 - cos ⁡ x - 2 sin ⁡ x ≤ 0 2 sin ⁡ x - 1 cos ⁡ x - 1 ≤ 0 Case I: cos ⁡ x = 1 ⇒ x = 0 Case II: otherwise cos ⁡ x 1 ⇒ 2 sin ⁡ x - 1 ≥ 0 ⇒ sin ⁡ x ≥ 1 2 ⇒ x ∈ π 6 , 5 π 6 Hence, from case I and case II, x ∈ π 6 , 5 π 6 ∪ 0

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