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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

The number of solutions for the equation ^ 3 xcosx + ^ 2 x ^ 2 x +...

The number of solutions for the equation sin⁡3xcosx+sin⁡2xcos⁡2x+sin⁡xcos⁡3x=1\sin^{3}xcosx + \sin^{2}x\cos^{2}x + \sin x\cos^{3}x = 1 in the interval [0,2π]\left[0, 2\pi\right] is/are

Options

  1. A00
  2. B22
  3. C33
  4. DInfinite

Correct answer

A. 00

Step-by-step solution

sin⁡3xcosx+sin⁡2xcos⁡2x+sin⁡xcos⁡3x=1\sin^{3}xcosx + \sin^{2}x\cos^{2}x + \sin x\cos^{3}x = 1

⇒sin⁡xcos⁡x(sin⁡2x+sin⁡xcos⁡x+cos⁡2x)=1⇒ \sin x \cos x\left(\sin^{2}x + \sin x \cos x + \cos^{2}x\right) = 1

⇒sin⁡2x2(1+sin⁡2x2)=1⇒ \dfrac{\sin 2x}{2}\left(1 + \dfrac{\sin 2x}{2}\right) = 1

⇒sin⁡2x(2+sin⁡2x)=4⇒ \sin 2x\left(2 + \sin 2x\right) = 4

⇒sin⁡22x+2sin⁡2x−4=0⇒ \sin^{2}2x + 2\sin 2x - 4 = 0

⇒sin⁡2x=−2±4+162⇒ \sin 2x = \dfrac{- 2 \pm \sqrt{4 + 16}}{2}

⇒sin⁡2x=−1±5⇒ \sin 2x = - 1 \pm \sqrt{5}

This is not possible. \left(Since - 1 \le \sin 2x \le 1\right)</p><p>Hencethegivenequationhasnosolution.</p></p><p>Hence the given equation has no solution.</p>

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