JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer
    Let \[S=\{\theta \in [-2\pi ,2\pi ]:2co{{s}^{2}}\theta +3sin\theta =0\}.\] Then the sum of the elements of S is                                                             [JEE Main 9-4-2019 Morning]

    A) \[\frac{13\pi }{6}\]                             

    B) \[\pi \]

    C) \[2\pi \]

    D) \[\frac{5\pi }{3}\]

    Correct Answer: C

    Solution :

    \[2\left( 1-{{\sin }^{2}}\theta  \right)+3\sin \theta =0\]           \[\Rightarrow 2{{\sin }^{2}}\theta -3\sin \theta -2=0\]           \[\Rightarrow (2\sin \theta +1)(\sin \theta -2)=0\]           \[\Rightarrow \sin \theta =-\frac{1}{2};\sin \theta =2(reject)\] roots\[\pi +\frac{\pi }{6},2\pi -\frac{\pi }{6},-\frac{\pi }{6},-\pi +\frac{\pi }{6}\] \[\Rightarrow \]sum of values\[=2\pi \]                       

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