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Exercise 1.3 · Q6

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

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Step 1. The function 11−2sin⁡x\dfrac1{1-2\sin x} is undefined exactly where the denominator vanishes: 1−2sin⁡x=0  ⟺  sin⁡x=121-2\sin x=0\iff\sin x=\dfrac12.

Step 2. The general solution of sin⁡x=12\sin x=\dfrac12 is x=nπ+(−1)nπ6, n∈Zx=n\pi+(-1)^n\dfrac\pi6,\ n\in Z (this single formula captures both families x=π6+2nπx=\dfrac\pi6+2n\pi and x=π−π6+2nπ=5π6+2nπx=\pi-\dfrac\pi6+2n\pi=\dfrac{5\pi}6+2n\pi). …

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