CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    If \[1+\sin +{{\sin }^{2}}x+...\] upto \[\infty =4+2\sqrt{3},\] \[0<x<\pi \]and \[x\ne \frac{\pi }{2},\] then x is equal to

    A)  \[\frac{\pi }{3},\frac{5\pi }{6}\]

    B)  \[\frac{2\pi }{3},\frac{\pi }{6}\]

    C)  \[\frac{\pi }{3},\frac{2\pi }{3}\]

    D)  \[\frac{\pi }{6},\frac{\pi }{3}\]

    Correct Answer: C

    Solution :

    Given,    \[1+\sin x+{{\sin }^{2}}x+....\infty =4+2\sqrt{3}\] \[\Rightarrow \]               \[\frac{1}{1-\sin x}=4+2\sqrt{3}\] \[\Rightarrow \]               \[1-\sin x=\frac{1}{4+2\sqrt{3}}\times \frac{4-2\sqrt{3}}{4-2\sqrt{3}}\] \[\Rightarrow \]               \[1-\sin x=\frac{4-2\sqrt{3}}{4}\] \[\Rightarrow \]               \[\sin x=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\] \[\Rightarrow \]               \[x=\frac{\pi }{3},\frac{2\pi }{3}\]


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