Verified Solution • 2025–26 Session moderate

QJEEM19Mad5ec: JEE Main 2019 Mathematics - Class 11 JEEM19Mad5ec Solution: Let S = { [–2 , 2 ] : 2cos 2 + 3sin = 0}. Then the sum of the…

Let S = { θ\theta ∈\in [–2 π\pi , 2 π\pi ] : 2cos2 θ\theta + 3sin θ\theta = 0}. Then the sum of the elements of S is

Multiple Choice Options

  • A π\pi
  • B 2 π\pi
  • C 13π6{{13\pi } \over 6}
  • D 5π3{{5\pi } \over 3}

Step-by-Step Verified Solution

2cos2 θ\theta + 3sin θ\theta = 0

⇒\Rightarrow 2(1 - sin2 θ\theta ) + 3sin θ\theta = 0

⇒\Rightarrow 2sin2 θ\theta - 3sin θ\theta - 2 = 0

⇒\Rightarrow 2sin2 θ\theta - 4sin θ\theta + sin θ\theta - 2 = 0

⇒\Rightarrow (2sin θ\theta + 1)(sin θ\theta - 2) = 0

∴\therefore sin θ\theta = 2 (Not possible)

sin θ\theta = −12- {1 \over 2}

∴\therefore θ\theta = n π\pi + (-1)n (−π6)\left( { - {\pi \over 6}} \right)

When n = 0, θ\theta = −π6{ - {\pi \over 6}}

When n = 1, θ\theta = 7π6{{7\pi } \over 6}

When n = -1, θ\theta = −5π6{-{5\pi } \over 6}

When n = 2, θ\theta = 11π6{{11\pi } \over 6}

∴\therefore Sum of all solutions in [–2 π\pi , 2 π\pi ] is

= −π6+7π6−5π6+11π6- {\pi \over 6} + {{7\pi } \over 6} - {{5\pi } \over 6} + {{11\pi } \over 6} = 2 π\pi

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