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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 solutions The equation 2cot⁡2x−3cot⁡3x=tan⁡2x2 \cot 2 x-3 \cot 3 x=\tan 2 x hasParagraph: In reducing a given equation to the standard forms ( x= α etc ) we apply several trigonometric or algebraicParagraph: In reducing a given equation to the standard forms ( x= α etc ) we apply several trigonometric or algebraic

Options

  1. Atwo solutions in \left(0, \frac{\pi}{3}\right)
  2. Bone solution in \left(0, \frac{\pi}{3}\right)
  3. Cno solution in (−∞,∞)(-\infty, \infty)
  4. Dnone of these

Correct answer

C. no solution in (−∞,∞)(-\infty, \infty)

Step-by-step solution

2cos⁡2xsin⁡2x−3cos⁡3xsin⁡3x=sin⁡2xcos⁡2x2(2cos⁡2x−1)2sin⁡xcos⁡x−34cos⁡3x−3cos⁡x3sin⁡x−4sin⁡3x=2sin⁡xcos⁡x2cos⁡2x−12 \frac{\cos 2 x}{\sin 2 x}-3 \frac{\cos 3 x}{\sin 3 x}=\frac{\sin 2 x}{\cos 2 x}2 \frac{\left(2 \cos ^2 x-1\right)}{2 \sin x \cos x}-3 \frac{4 \cos ^3 x-3 \cos x}{3 \sin x-4 \sin ^3 x}=\frac{2 \sin x \cos x}{2 \cos ^2 x-1} Now Let cos⁡2x=t\cos ^2 \mathrm{x}=\mathrm{t} in above i.e. t∈(0,1)−{12}t \in(0,1)-\left\{\frac{1}{2}\right\} Now we have 2t−1−3t4t−34t−1=2t(1−t)2t−12 t-1-3 t \frac{4 t-3}{4 t-1}=\frac{2 t(1-t)}{2 t-1} Solving t=±1\mathrm{t}= \pm 1 but t∈(0,1)−{12}\mathrm{t} \in(0,1)-\left\{\frac{1}{2}\right\} So there is no tt, so there is no xx Paragraph: In reducing a given equation to the standard forms ( x= α etc ) we apply several trigonometric or algebraic
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