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Need clarity, kindly explain! - Trigonometry - JEE Main-5

Let S=\left \{ \theta \epsilon \left [ -2\pi ,2\pi \right ] :2cos^{2}\theta +3sin\theta =0\right \}

Then the sum of the elements of S is :

  • Option 1)

     \frac{13\pi }{6}          

  • Option 2)

    \frac{5\pi }{3}

  • Option 3)

    2\pi

  • Option 4)

    \pi

 
Answers (1)
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Given equation 

   2cos^{2}\theta +3sin\theta =0

2-2sin^{2}\theta +3sin \theta =0

OR

2sin^{2}\theta -3sin \theta-2 =0

sin\theta =-\frac{1}{2}\; or\; 2

sin\theta \neq 2

Hence solution is \left [ -2\pi ,2\pi \right ]

-\pi +\frac{\pi }{6},\frac{-\pi }{6},\pi +\frac{\pi }{6},2\pi -\frac{\pi }{6}

Sum=2\pi


Option 1)

 \frac{13\pi }{6}          

Option 2)

\frac{5\pi }{3}

Option 3)

2\pi

Option 4)

\pi

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