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March 25, 2017

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Find all solutions of equation on interval [0,2pi]

1=cot^2x + cscx

  • Trig - ,

    Multiply both sides by sin^2(x).
    sin^2(x)=cos^2(x)+sin(x)
    sin^2(x)=1-sin^2(x)+sin(x)
    2sin^2(x)-sin(x)-1=0
    (2sin(x)+1)(sin(x)-1)=0
    sin(x)=-1/2, x=7pi/6, x=11pi/6
    or
    sin(x)=1, x=pi/2

  • Trig - ,

    Thank you so much!

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