JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Solution of trigonometrical equations

  • question_answer
    The solution of equation \[{{\cos }^{2}}\theta +\sin \theta +1=0\] lies in the interval       [UPSEAT 2004; IIT 1992]

    A) \[\left( -\frac{\pi }{4},\frac{\pi }{4} \right)\]

    B) \[\left( \frac{\pi }{4},\frac{3\pi }{4} \right)\]

    C) \[\left( \frac{3\pi }{4},\frac{5\pi }{4} \right)\]

    D) \[\left( \frac{5\pi }{4},\frac{7\pi }{4} \right)\]

    Correct Answer: D

    Solution :

    We have, \[{{\cos }^{2}}\theta +\sin \theta +1=0\] Þ  \[1-{{\sin }^{2}}\theta +\sin \theta +1=0\] Þ  \[{{\sin }^{2}}\theta -\sin \theta -2=0\] Þ \[(\sin \theta +1)\,(\sin \theta -2)=0\] \[\sin \theta =2\], which is not possible and\[\sin \theta =-1\]. Therefore, solution of given equation lies in the interval\[\left( \frac{5\pi }{4},\,\frac{7\pi }{4} \right)\].


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