# 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°

5π/6 + πk

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