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

Question

The solution set of x∈ (- π , π ) for the inequality sin 2 x + 1≤ cos ⁡ x+2sin ⁡ x is (A) x∈ [0 , (π /6)] (B) x∈ [(π /6) , (5 π /6)]∪ 0  (C) x∈ [- (π /6) , (5 π /6)] (D) x∈ [- (π /2) , (π /2)]

Tardigrade Admin · Accepted Answer

2sin xcos â¡ x+1-cos â¡ x-2sin â¡ x≤ 0 (2 sin x - 1)(cos â¡ x - 1)≤ 0 Case I: cos x=1⇒ x=0 Case II: otherwise cos x < 1 ⇒ 2sin x-1≥ 0⇒ sin â¡ x≥ (1/2) ⇒ x∈ [(π /6) , (5 π /6)] Hence, from case I and case II, x∈ [(π /6) , (5 π /6)]∪ 0

Q. The solution set of $x \in (- π, π)$ for the inequality $s in 2 x + 1 \leq cos x + 2 s in x$ is

$x \in [0, \frac{π}{6}]$

$x \in [\frac{π}{6}, \frac{5 π}{6}] \cup {0}$

$x \in [- \frac{π}{6}, \frac{5 π}{6}]$

$x \in [- \frac{π}{2}, \frac{π}{2}]$

Q. The solution set of x∈(−π,π) for the inequality sin2x+1≤cos⁡x+2sin⁡x is

x∈[0,6π​]

x∈[6π​,65π​]∪{0}

x∈[−6π​,65π​]

x∈[−2π​,2π​]

2sinxcos⁡x+1−cos⁡x−2sin⁡x≤0 (2sinx−1)(cos⁡x−1)≤0 Case I: cosx=1⇒x=0 Case II: otherwise cosx<1 ⇒2sinx−1≥0⇒sin⁡x≥21​ ⇒x∈[6π​,65π​] Hence, from case I and case II, x∈[6π​,65π​]∪{0}

Q. The solution set of $x \in (- π, π)$ for the inequality $s in 2 x + 1 \leq cos x + 2 s in x$ is

$x \in [0, \frac{π}{6}]$

$x \in [\frac{π}{6}, \frac{5 π}{6}] \cup {0}$

$x \in [- \frac{π}{6}, \frac{5 π}{6}]$

$x \in [- \frac{π}{2}, \frac{π}{2}]$