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March 24, 2017

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Can someone please help me with this?
Find all solutions to the equation in the interval [0,2pi)
cos2x=sinx

I know I have to use some sort of identity, but I have no idea how to go about to solve this.

  • Trig. - ,

    Yes, by making use of the identity
    cos2x = cos²x-sin²x=1-2sin²x, the given equation can be transformed to:
    2sin²x+sinx-1=0
    which is a quadratic in sin(x).
    Solving the quadratic by factoring,
    (2sinx-1)(sinx+1)=0
    or sin(x)=1/2 or sin(x)=-1
    Find all roots in the interval [0,2π].

  • Trig. - ,

    Thank you very much for the help

  • Trig. :) - ,

    You are welcome!

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