If x,y∈0,15, then the number of solutions x,y of the equation  3cosec2x−1×4y2−4y+2≤1 is    

If x,y0,15, then the number of solutions x,y of the equation  3cosec2x1×4y24y+21 is    

  1. A

    13

  2. B

    17

  3. C

    15

  4. D

    5

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    Solution:

    We have 3cot2x2y12+11.

    But 3cot2x1and 2y121

    3cot2x=1and 2y12+1=1

    cot2x=0,y=12

    x=π2,3π2,5π2,7π2,9π2   x0,15

     

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