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J & K CET Engineering J and K - CET Engineering Solved Paper-2008

  • question_answer
    The number of solutions of \[\sin x=\sin 2x\]between \[\frac{-\pi }{2}\] and \[\frac{\pi }{2}\] is

    A) \[3\]               

    B) \[2\]

    C) \[1\]                

    D) \[0\]

    Correct Answer: A

    Solution :

    Given,   \[\sin x=\sin 2x\] \[\Rightarrow \] \[\sin \,x=2\,\sin x\,\cos x\] \[\Rightarrow \] \[\sin x(1-2\,\cos x)=0\] \[\Rightarrow \] \[\sin x=0\] or \[\cos \,x=\frac{1}{2}\] \[\Rightarrow \] \[x={{0}^{o}}\] or \[x={{60}^{o,}}\,-{{60}^{o}}\] Hence, number of solution is 3.


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