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JEE Main 202122 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
22 July
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2021 · 22 July · Morning Shift · Mathematics · Trigonometric Equations

The number of solutions of ^ 7 x + ^ 7 x = 1, x ∈ [ 0, 4π ] is...

The number of solutions of sin⁡7x+cos⁡7x=1,x∈[0,4π]\sin^{7}x + \cos^{7}x = 1, x \in [ 0, 4\pi ] is equal to

Options

  1. A1111
  2. B77
  3. C55
  4. D99

Correct answer

C. 55

Step-by-step solution

We know that 0≤sin⁡2x≤10 \le \sin^{2}x \le 1 and 0≤cos⁡2x≤10 \le \cos^{2}x \le 1

Also, we know for a number which lie between 00 to 1,1, higher the power of the number the less is its value.

⇒sin⁡7x≤sin⁡2x≤1…(1)⇒ \sin^{7}x \le \sin^{2}x \le 1 \ldots \left(1\right)

and cos⁡7x≤cos⁡2x≤1…(2)\cos^{7}x \le \cos^{2}x \le 1 \ldots \left(2\right)

Also, we know that sin⁡2x+cos⁡2x=1\sin^{2}x + \cos^{2}x = 1

This means the equality must hold for \left(1\right)and(2)and \left(2\right)

⇒sin⁡7x=sin⁡2x⇒ \sin^{7}x = \sin^{2}x and cos⁡7x=cos⁡2x\cos^{7}x = \cos^{2}x

⇒sin⁡2x(sin⁡5x−1)=0⇒ \sin^{2}x\left(\sin^{5}x - 1\right) = 0 and cos⁡2x(cos⁡5x−1)=0\cos^{2}x\left(\cos^{5}x - 1\right) = 0

⇒sin⁡x=0⇒ \sin x = 0 or sin⁡x=1\sin x = 1 and cos⁡x=0\cos x = 0 or cos⁡x=1\cos x = 1⇒x=0,2π,4π,π2,5π2.⇒ x = 0, 2\pi, 4\pi, \dfrac{\pi}{2}, \dfrac{5\pi}{2} .

Thus, there are total 55 solutions.

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