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JEE Advanced 2014Paper 2
Exam
JEE Advanced
Year
2014
Paper
Paper 2
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Advanced 2014 · Paper 2 · Mathematics · Trigonometric Equations

For xϵ(0, π), the equation sin x + 2 2x - 3x = 3 has

For xϵ(0,π),xϵ\left(0, \pi\right), the equation sinx+2sin⁡2x−sin⁡3x=3sin x + 2\sin 2x - \sin 3x = 3 has

Options

  1. AInfinitely solution
  2. BThree solutions
  3. COne solution
  4. DNo solution

Correct answer

D. No solution

Step-by-step solution

sinx+2sin⁡2x−sin⁡3x=3sin x + 2\sin 2x - \sin 3x = 3
sinx+4sinxcosx−3sinx+4sin⁡3x=3sin x + 4sin xcos x - 3sin x + 4\sin^{3} x = 3
sinx[−2+4cosx+4(1−cos⁡2x)]=3sin x\left[- 2 + 4cos x + 4\left(1 - \cos^{2} x\right)\right] = 3
sinx[2−(4cos⁡2x−4cosx+1)+1]=3sin x\left[2 - \left(4\cos^{2} x - 4cos x + 1\right) + 1\right] = 3
sinx[3−(2cosx−1)2]=3sin x{\left[3 - {\left(2cos x - 1\right)}^{2}\right]}^{} = 3
⇒sinx=1and2cosx−1=0⇒ sin x = 1and2cos x - 1 = 0
⇒x=π2andx=π3⇒ x = \dfrac{\pi}{2}andx = \dfrac{\pi}{3}
Which is not possible at same time
Hence, no solution
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