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Question 148 of 175

Q.Find the domain of 11−2sin⁡x\dfrac{1}{1-2\sin x}.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2020Subjective· 3mImportance★★★★★
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Exclude every xx where sin⁡x=12\sin x=\dfrac12, since that makes the denominator zero.

For 11−2sin⁡x\dfrac{1}{1-2\sin x} to be defined, we need the denominator to be nonzero:

1−2sin⁡xe0 ⇒ sin⁡xe12.1-2\sin x e0\ \Rightarrow\ \sin x e\dfrac12.

The general solution of sin⁡x=12\sin x=\dfrac12 is x=nπ+(−1)nπ6x=n\pi+(-1)^n\dfrac\pi6 for n∈Zn\in\mathbf Z — concretely this gives x=…,π6,5π6,π6+2π,5π6+2π,…x=\ldots,\dfrac\pi6,\dfrac{5\pi}6,\dfrac\pi6+2\pi,\dfrac{5\pi}6+2\pi,\ldots (i.e. π6\dfrac\pi6 and 5π6\dfrac{5\pi}6 together with all their coterminal angles obtained by adding multiples of 2π2\pi).

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