Domain of sin−1(2x)+π6 is

f(x)=sin−12x+π6 is defined iffsin−12x+π6≥0⇒ sin−12x≥−π6⇒ 2x≥sin(−π6)⇒ 2x≥−12⇒x≥−14But Domain of sin−1x is [−1, 1]∴ −1≤2x≤1⇒ −12≤x≤12∴ x∈ [−14, 12]

Domain of sin−1(2x)+π6 is

f(x)=sin−12x+π6 is defined iffsin−12x+π6≥0⇒ sin−12x≥−π6⇒ 2x≥sin(−π6)⇒ 2x≥−12⇒x≥−14But Domain of sin−1x is [−1, 1]∴ −1≤2x≤1⇒ −12≤x≤12∴ x∈ [−14, 12]

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$D o m a i n o f \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}} i s$

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$[- \frac{1}{4}, \frac{1}{2}]$

$[- \frac{1}{2}, \frac{1}{2}]$

$(- \frac{1}{2}, \frac{1}{9})$

$[- \frac{1}{4}, \frac{1}{4}]$

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{\sin^{- 1} 2 x + \frac{π}{6}} i s d e f i n e d i f f \\ \sin^{- 1} 2 x + \frac{π}{6} \geq 0 \\ \Rightarrow \sin^{- 1} 2 x \geq \frac{- π}{6} \\ \Rightarrow 2 x \geq \sin (\frac{- π}{6}) \\ \Rightarrow 2 x \geq \frac{- 1}{2} \\ \Rightarrow x \geq \frac{- 1}{4} \end{array}$

$\begin{array}{l} B u t D o m a i n o f \sin^{- 1} x i s [- 1, 1] \\ ∴ - 1 \leq 2 x \leq 1 \\ \Rightarrow \frac{- 1}{2} \leq x \leq \frac{1}{2} \\ ∴ x \in [\frac{- 1}{4}, \frac{1}{2}] \end{array}$