Range of the function f(x) = frac{1}{2 - sin | vedclass

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(B) Given the function $f(x) = \frac{1}{2 - \sin 3x}$.We know that the range of $\sin 3x$ is $[-1, 1]$.Let $u = \sin 3x$,then $u \in [-1, 1]$.The func

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$[\frac{1}{3}, 1]$

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(B) Given the function $f(x) = \frac{1}{2 - \sin 3x}$.We know that the range of $\sin 3x$ is $[-1, 1]$.Let $u = \sin 3x$,then $u \in [-1, 1]$.The function becomes $f(u) = \frac{1}{2 - u}$.To find the range,we evaluate the function at the boundaries of $u$:When $u = -1$,$f(-1) = \frac{1}{2 - (-1)} = \frac{1}{3}$.When $u = 1$,$f(1) = \frac{1}{2 - 1} = 1$.Since the function is continuous and monotonic within this interval,the range of the function is $[\frac{1}{3}, 1]$.

Range of the function $f(x) = \frac{1}{2 - \sin 3x}$ is

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