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find domain and range:

f(x) = 1/2-sin 3x

$Given f(x)= \frac{1}{2 - \sin 3 x} l e t y = \frac{1}{2 - \sin 3 x} - - - - (i) Domain of f(x) is (- \inf inity,infinity) From (i) 2 - \sin 3 x = \frac{1}{y} \Rightarrow \sin 3 x = \frac{2 y - 1}{y} \Rightarrow x = \frac{1}{3} \sin^{- 1} (\frac{2 y - 1}{y}) (∵ y = \frac{1}{2 - \sin 3 x} ∴ y > 0 as - 1 \leq \sin 3 x \leq 1) For x to be real - 1 \leq \frac{2 y - 1}{y} \leq 1 \Rightarrow - y \leq 2 y - 1 \leq y (∵ y > 0) \Rightarrow 2 y - 1 \geq - y and 2 y - 1 \leq y \Rightarrow y \geq \frac{1}{3} and y \leq 1 ∴ Range of f(x)= [\frac{1}{3}, 1]$

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