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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

For the equation 2( x- 2 x)- 2 x(1+2 x)+2 x=0, x ∈[0,2 π]-

For the equation 2(sin⁡x−cos⁡2x)−sin⁡2x(1+2sin⁡x)+2cos⁡x=0,x∈[0,2π]−2(\sin x-\cos 2 x)-\sin 2 x(1+2 \sin x)+2 \cos x=0, x \in[0,2 \pi]-

Options

  1. Athe values of xx satisfying the given equation forms an A.P.
  2. Bthe sum of all possible values of xx satisfying the given equation is 4π4 \pi
  3. Cthe sum of all possible values of x in [0,2π][0,2 \pi] satisfying the given equation is 9π2\frac{9 \pi}{2}
  4. Dthere are five solutions of the equation in [0,2π][0,2 \pi]

Correct answer

D. there are five solutions of the equation in [0,2π][0,2 \pi]

Step-by-step solution

2sin⁡x−2cos⁡2x−sin⁡2x−2sin⁡xsin⁡2x+2cos⁡x=02sin⁡x−2cos⁡x−sin⁡2x−cos⁡x+cos⁡3x+2cos⁡x=02sin⁡x(1−cos⁡x)+4cos⁡3x−3cos⁡x−2(2cos⁡2x−1)+cos⁡x=02sin⁡x(1−cos⁡x)+4cos⁡3x−4cos⁡2x−2cos⁡x+2=02sin⁡x(1−cos⁡x)+4cos⁡2x(cos⁡x−1)−2(cos⁡x−1)=0(1−cos⁡x){2sin⁡x+2−4(1−sin⁡2x)}=0(1−cos⁡x){4sin⁡2x+2sin⁡x−2}=0(1−cos⁡x)(sin⁡x+1)(2sin⁡x−1)=0x=0,2π,3π2,π6,5π6\begin{aligned} & 2 \sin x-2 \cos 2 x-\sin 2 x-2 \sin x \sin 2 x+2 \cos x=0 \\ & 2 \sin x-2 \cos x-\sin 2 x-\cos x+\cos 3 x+2 \cos x=0 \\ & 2 \sin x(1-\cos x)+4 \cos ^3 x-3 \cos x \\ & -2\left(2 \cos ^2 x-1\right)+\cos x=0 \\ & 2 \sin x(1-\cos x)+4 \cos ^3 x-4 \cos ^2 x-2 \cos x+2=0 \\ & 2 \sin x(1-\cos x)+4 \cos ^2 x(\cos x-1)-2(\cos x-1)=0 \\ & (1-\cos x)\left\{2 \sin x+2-4\left(1-\sin ^2 x\right)\right\}=0 \\ & (1-\cos x)\left\{4 \sin ^2 x+2 \sin x-2\right\}=0 \\ & (1-\cos x)(\sin x+1)(2 \sin x-1)=0 \\ & x=0,2 \pi, \frac{3 \pi}{2}, \frac{\pi}{6}, \frac{5 \pi}{6}\end{aligned}
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