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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The sum of roots of the equation ^ 4 x - ^ 2 xsinx + 2 ^ 2 x + x =...

The sum of roots of the equation sin⁡4x−cos⁡2xsinx+2sin⁡2x+sin⁡x=0\sin^{4}x - \cos^{2}xsinx + 2\sin^{2}x + \sin x = 0 for 0≤x≤3π0 \le x \le 3\pi is equal to

Options

  1. A3π4\dfrac{3\pi}{4}
  2. B2π2\pi
  3. C4π4\pi
  4. D6π6\pi

Correct answer

D. 6π6\pi

Step-by-step solution

sin⁡x(sin⁡3x−cos⁡2x+2sin⁡x+1)=0\sin x\left(\sin^{3}x - \cos^{2}x + 2\sin x + 1\right) = 0
sin⁡x(sin⁡3x+2sin⁡x+1−cos⁡2x)=0\sin x\left(\sin^{3}x + 2\sin x + 1 - \cos^{2}x\right) = 0
sin⁡x(sin⁡3x+sin⁡2x+2sin⁡x)=0\sin x\left(\sin^{3}x + \sin^{2}x + 2\sin x\right) = 0
sin⁡2x(sin⁡2x+sin⁡x+2)=0\sin^{2}x\left(\sin^{2}x + \sin x + 2\right) = 0
⇒sin⁡x=0⇒ \sin x = 0 as sin⁡2x+sin⁡x+2≠0\sin^{2}x + \sin x + 2 \ne 0
⇒x=nπ,n∈I⇒ x = n\pi, n \in I
x∈[0,3π],x=0,π,2π,3πx \in \left[0,3\pi\right], x = 0, \pi, 2\pi, 3\pi
⇒⇒ sum of roots =6π= 6\pi
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