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Find all solutions on the interval (0, 2pi) of the equation: 2(sin^2)t-3sint+1=0
how do you get this one started?

  • Math -

    let s=sin(t), so
    2(sin^2)t-3sint+1=0
    becomes
    2s²-3s+1=0
    (2s-1)(s-1)=0
    =>
    s=1/2 or s=1
    =>
    sin(x)=1/2 or sin(x)=1
    Now solve for x for 0≤x≤2π

  • Math -

    That makes so much sense. Thank you!

  • Math :) -

    You're welcome!

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