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JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    Sum of all the solutions of the equation \[\cos 2x+{{\sin }^{2}}2x=1\], which lie in the interval \[[0,2\pi ]\] is equal to

    A) \[4\pi \]

    B) \[5\pi \]

    C) \[7\pi \]

    D) \[8\pi \]

    Correct Answer: C

    Solution :

    \[\cos 2x(1-\cos 2x)=0\] \[\cos 2x=1\]     or            \[\cos 2x=0\] \[\therefore \]\[x=n\pi \]            or            \[x=(2n-1)\frac{\pi }{4}\] \[0,\,\,\pi ,\,\,2\pi \]       or            \[\frac{\pi }{4},\,\,\frac{3\pi }{4},\,\,\frac{5\pi }{4},\,\,\frac{7\pi }{4}=7\pi \]


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