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

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 = 18030 = 150° or Ø = 36030 = 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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