4tanθ + √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
find all solutions to 4tanθ + √3 = tanθ
1 answer
1 answer