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Posted by on Wednesday, February 22, 2012 at 10:25pm.

Solve over the indicated interval:
2sin^2x=sqrt2sinx , [-180,180)

don't how to get x by itself

please help and thank you

  • Trig Help (URGENT) - , Wednesday, February 22, 2012 at 10:57pm

    let sinx = y, so you get

    2y^2 = √(2y)
    square both sides
    4y^4 = 2y
    4y^4 - 2y = 0
    2y(2y^3 - 1) = 0
    y = 0 or y^3 = 1/2 --- y = (.5)^(1/3) = appr .7937

    if y = 0, then sinx = 0
    x = -180°, 0° , 180°

    if y = .7937 , then sin x = + .7937 (x must be in I or II)
    x = 52.53 or 180-52.53 = 127.47°

    but we squared, so our answers MUST be verfied

    if x = 180 or -180
    LS = 0 , RS = 0 , ✔
    if x = 0, ✔
    if x = 52.53°
    LS = 1.2599 , RS = 1.2599 ✔
    if x = 127.47
    Ls = 1.2599 , RS = 1.2599 ✔

    All 5 answers above are correct

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