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
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
1 answer
1 answer