Answer:
θ = 5π/6 rad and 11π/6 rad
Explanation:
Given the expression cotθ+√3=0
Subtract √3 from both sides
cotθ+√3-√3=0-√3
cotθ = -√3
Since cotθ = 1/tanθ
1/tanθ = -√3
Reciprocate both sides:
tanθ = -1/√3
θ = tan^-1(-1/√3)
θ = -30°
Since the angle is negative, and tanθ is negative in the second and fourth quadrant.
In the second quadrant;
θ = 180-30
θ = 150°
Since 180° = πrad
150° = 150π/180
150° = 5π/6 rad
In the fourth quadrant;
θ = 360-30
θ = 330°
Since 180° = πrad
330° = 330π/180
330° = 11π/6 rad
Hence the solutions are 5π/6 rad and 11π/6 rad.