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April 18, 2014

April 18, 2014

Posted by **Ama** on Tuesday, November 27, 2012 at 3:33pm.

tan θ = - root of 3/3

- precalculus -
**Reiny**, Tuesday, November 27, 2012 at 5:03pmtan 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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