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March 26, 2017

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Approximate the equation's soultions in the interval (o, 2pi). If possible find the exact solutions.

sin 2x sinx = cosx

I do not know where to start.

  • Pre-Cal - ,

    Expand sin 2x using the identity:
    sin2x=2sin(x)cos(x)
    We'll get:
    2cos(x)sin²(x)=cos(x)
    Transpose left-hand-side to the right:
    2cos(x)((1/2)-sin²(x))=0
    So
    cos(x)=0, or
    sin²(x)=1/2
    Solve for x in (0,2π) as specified.

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