trigoo

(sinx-cosx)^2=sin2x

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  1. What are we doing? Solving?

    of so, then square both sides

    sin^2 - 2sinxcosx + cos^2 = sin2x
    1 - sin2x = sin2x
    2sin2x = 1
    sin2x = 1/2
    2x = 30° or 150°
    x = 15° or 75°

    since we squared, all answers must be checked ..
    if x=15°
    LS = (sin15 - cos15)^2 = .5
    RS = sin 30 = .5 = Ls, so x=15° works

    if x = 75°
    Ls = (sin75-cos75)^2 = .5
    RS = sin 150 = .5 , so x = 75° works

    In radians your solution would be
    x = π/12 and 5π/12

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  2. thnx thnx so much

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