The solution set of x ∈ (- π, π) for the inequality 2x + 1 ≤ cos x + 2sin x is

Correct answer: B. x ∈ [(π)/(6), (5π)/(6)] \ 0\ 2sin xcos x + 1 - cos x - 2sin x ≤ 0 (2sin x - 1)(cos x - 1)≤ 0 Case I: cos x = 1 ⇒ x = 0 Case II: otherwise cos x ⇒ 2sin x - 1 ≥ 0 ⇒ sin x ≥ (1)/(2) ⇒ x ∈ [(π)/(6), (5π)/(6)] Hence, from case I and case II, x ∈ [(π)/(6), (5π)/(6)] \ 0\

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The solution set of x ∈ (- π, π) for the inequality 2x + 1 ≤ cos x...

The solution set of

x \in

\left(- \pi, \pi\right)

for the inequality

\sin 2x + 1 ≤ cos x + 2sin x is

Options

A $x \in \left[0, \dfrac{\pi}{6}\right]$
B $x \in \left[\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right] \cup \left\{0\right\}$
C $x \in \left[- \dfrac{\pi}{6}, \dfrac{5\pi}{6}\right]$
D $x \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Correct answer

B. $x \in \left[\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right] \cup \left\{0\right\}$

Step-by-step solution

2sin xcos x + 1 - cos x - 2sin x \le 0

<br />\left(2sin x - 1\right)\left(cos x - 1\right)\le 0

Case I:

cos x = 1 ⇒ x = 0

Case II: otherwise

cos x < 1

⇒ 2sin x - 1 \ge 0 ⇒ sin x \ge \dfrac{1}{2}

⇒ x \in \left[\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right]<

br />Hence, from case I and case II,

x \in \left[\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right]\cup \left\{0\right\}

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