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September 30, 2014

September 30, 2014

Posted by **isla** on Wednesday, March 20, 2013 at 10:12pm.

- math -
**Reiny**, Wednesday, March 20, 2013 at 10:48pm4tanθ + √3 = tanθ

3tanØ = -√3

tanØ = -√3/3

Ø must be in II or IV

Ø = 150° or 330°

or

Ø = 5π/6 or 11π/6

period of tanØ = 90°

general solutions:

=(150+90k)° or 330° +90k°

do the same with the radian answer by adding πk/2 to each of the answers

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