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precalculus

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Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.)
tan θ = - root of 3/3

  • precalculus - ,

    tan Ø = -√3/3
    so Ø must be in quadrants II or IV
    angle in standard position = 30° ( take tan^-1 (+√3/3) )

    so Ø = 180-30 = 150° or Ø = 360-30 = 330°

    since the period of the tangent curve is 180°

    the general solution is
    150° + 180k°

    or in radians:
    5π/6 + πk

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