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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Paragraph: In reducing a given equation to the standard forms ( x=...

Paragraph: In reducing a given equation to the standard forms (sin⁡x=sin⁡α(\sin x=\sin \alpha etc )) we apply several trigonometric or algebraic transformations. As a result of which the canonical form finally obtained may not be equivalent to the original equation resulting either in loss of solutions or in the appearance of fake solutionsQuestion: The solution set of the equation 5−2sin⁡x=6sin⁡x−1\sqrt{5-2 \sin x}=6 \sin x-1 is

Options

  1. Anπ+(−1)nsin⁡−1(−29)or\mathrm{n} \pi+(-1)^n \sin ^{-1} \left(-\frac{2}{9}\right) or\mathrm{x}=\mathrm{n}π+(−\pi+(-1)^n \frac{\pi}{6}
  2. Bnπ+(−1)nπ6n \pi+(-1)^n \frac{\pi}{6}
  3. Cn π+(−\pi+(-1)^n \sin ^{-1} \left(\frac{2}{9}\right)
  4. DNull set

Correct answer

B. nπ+(−1)nπ6n \pi+(-1)^n \frac{\pi}{6}

Step-by-step solution

Let sin⁡\sin\mathrm{x}=\mathrm{t};t∈[−1,1]5−2t=6t−1≥0; \mathrm{t} \in[-1,1]\sqrt{5-2 t}=6 t-1 \geq 0 \quad i.e. t≥16\quad t \geq \frac{1}{6} So, t∈t \in\left[\frac{1}{6},1\right]Nowsquaring,Now squaring,\sqrt{5-2 \mathrm{t}}=(6 \mathrm{t}-1), \mathrm{t}=\frac{1}{2},-\frac{2}{9}, \mathrm{t}=-\frac{2}{9}not possible \left(\because \mathrm{t} \in \left[\frac{1}{6}, 1\right]\right)So,So,t=\frac{1}{2}, \sin x=\frac{1}{2}=\sin \frac{\pi}{6}, x=n \pi+(-1)^n \frac{\pi}{6}
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