If -1-sin x-1-sin x-23- then x lies in

|1 |sin x|1+|sin x||≥23 ⇒1|1+|sin x||≥23 Since 1+|sin x|≥1 ∀ x∈ℝ, ∴ nbsp;11+|sin x|≥23 ⇒ |sin x|≤12 From the graph, nbsp; |sin x|≤12 ⇒ x∈[ π6,π6] ∵ Perio ...

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Byju's Answer

Standard XIII

Mathematics

General Solution of Trigonometric Equation

If |1-|sin x|...

Question

If $∣ ∣ ∣ 1 - \frac{| sin x |}{1 + | sin x |} ∣ ∣ ∣ \geq \frac{2}{3},$ then $x$ lies in

[- \frac{π}{6}, \frac{π}{6}]

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[\frac{5 π}{6}, \frac{7 π}{6}]

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[- \frac{7 π}{6}, - \frac{5 π}{6}]

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[\frac{π}{6}, \frac{5 π}{6}]

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Solution

The correct option is C $[- \frac{7 π}{6}, - \frac{5 π}{6}]$

$∣ ∣ ∣ 1 - \frac{| sin x |}{1 + | sin x |} ∣ ∣ ∣ \geq \frac{2}{3} \Rightarrow \frac{1}{| 1 + | sin x | |} \geq \frac{2}{3}$
Since $1 + | sin x | \geq 1 \forall x \in R,$
$∴ \frac{1}{1 + | sin x |} \geq \frac{2}{3} \Rightarrow | sin x | \leq \frac{1}{2}$

From the graph,
$| sin x | \leq \frac{1}{2} \Rightarrow x \in [- \frac{π}{6}, \frac{π}{6}]$
$∵$ Period of $y = | sin x |$ is $π,$
$∴ x \in ⋃ n \in Z [n π - \frac{π}{6}, n π + \frac{π}{6}]$

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If ∣∣∣1−|sinx|1+|sinx|∣∣∣≥23, then x lies in

The correct option is C [−7π6,−5π6] ∣∣∣1−|sinx|1+|sinx|∣∣∣≥23⇒1|1+|sinx||≥23 Since 1+|sinx|≥1 ∀ x∈R, ∴11+|sinx|≥23⇒|sinx|≤12 From the graph, |sinx|≤12⇒x∈[−π6,π6] ∵ Period of y=|sinx| is π, ∴x∈⋃n∈Z[nπ−π6,nπ+π6]

If $∣ ∣ ∣ 1 - \frac{| sin x |}{1 + | sin x |} ∣ ∣ ∣ \geq \frac{2}{3},$ then $x$ lies in