Verified Solution • 2025–26 Session moderate

QJEEM21Ma1b68: JEE Main 2021 Mathematics - Class 11 JEEM21Ma1b68 Solution: The sum of all values of x in [0, 2 ], for which sin x + sin 2x …

The sum of all values of x in [0, 2 π\pi ], for which sin x + sin 2x + sin 3x + sin 4x = 0, is equal to :

Multiple Choice Options

  • A 8 π\pi
  • B 11 π\pi
  • C 12 π\pi
  • D 9 π\pi

Step-by-Step Verified Solution

(sin⁡x+sin⁡4x)+(sin⁡2x+sin⁡3x)=0(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0

⇒2sin⁡5x2{cos⁡3x2+cos⁡x2}=0\Rightarrow 2\sin {{5x} \over 2}\left\{ {\cos {{3x} \over 2} + \cos {x \over 2}} \right\} = 0

⇒2sin⁡5x2{2cos⁡xcos⁡x2}=0\Rightarrow 2\sin {{5x} \over 2}\left\{ {2\cos x\cos {x \over 2}} \right\} = 0

2sin⁡5x2=0⇒5x2=0,π,2π,3π,4π,5π2\sin {{5x} \over 2} = 0 \Rightarrow {{5x} \over 2} = 0,\pi ,2\pi ,3\pi ,4\pi ,5\pi

⇒x=0,2π5,4π5,6π5,8π5,2π\Rightarrow x = 0,{{2\pi } \over 5},{{4\pi } \over 5},{{6\pi } \over 5},{{8\pi } \over 5},2\pi

cos⁡x2=0⇒x2=π2⇒x=π\cos {x \over 2} = 0 \Rightarrow {x \over 2} = {\pi \over 2} \Rightarrow x = \pi

cos⁡x=0⇒x=π2,3π2\cos x = 0 \Rightarrow x = {\pi \over 2},{{3\pi } \over 2}

So, sum =6π+π+2π=9π= 6\pi + \pi + 2\pi = 9\pi

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