The domain of f(x)=1|sinx|+sinx is

f(x)=1|sinx|+sinxis defined iff |sinx|+sinx≠0|sinx|+sinx={2 sinxif 2nπ<x<(2n+1)π0Otherwise∴ Domain =∪n∈Z(2nπ, (2n+1)π)

The domain of f(x)=1|sinx|+sinx is

f(x)=1|sinx|+sinxis defined iff |sinx|+sinx≠0|sinx|+sinx={2 sinxif 2nπ<x<(2n+1)π0Otherwise∴ Domain =∪n∈Z(2nπ, (2n+1)π)

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$T h e d o m a i n o f f (x) = \frac{1}{| \sin x | + \sin x} i s$

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$\underset{n \in Z}{U} ((2 n + 1) π, 2 (n + 1) π)$

$ϕ$

$\underset{n \in Z}{U} (2 n π, (2 n + 1) π)$

answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{1}{| \sin x | + \sin x} \\ i s d e f i n e d i f f | \sin x | + \sin x \neq 0 \\ | \sin x | + \sin x = {\begin{cases} 2 \sin x i f 2 n π < x < (2 n + 1) π \\ 0 O t h e r w i s e \end{cases} \\ ∴ D o m a i n = \underset{n \in Z}{\cup} (2 n π, (2 n + 1) π) \end{array}$