I need help with this trig problem
cotθ=-√3/3 and 3/2π≤θ≤π, find θ
To find the value of θ when cotθ = -√3/3 and 3/2π ≤ θ ≤ π, we can use the inverse cotangent function (also known as arccotangent) to solve for θ.
The inverse cotangent of a given value, in this case, -√3/3, can be found using a calculator or a trigonometric table. The inverse cotangent function, arccot(x), returns the angle whose cotangent is equal to x.
arccot(-√3/3) ≈ 5π/6
However, we need to ensure that the angle θ lies between 3/2π and π, as given in the problem statement. The angle 5π/6 is in the first quadrant (0 to π/2), so it doesn't fall within the specified range.
To find another angle that matches the given conditions, we can consider the symmetry of the cotangent function. Since cot(x) = 1/tan(x), we know that the cotangent of an angle is positive in the second and fourth quadrants.
In the second quadrant, the reference angle (the acute angle formed between the terminal side of θ and the x-axis) whose cotangent is -√3/3 is π - arccot(-√3/3).
π - arccot(-√3/3) ≈ π - (5π/6) ≈ π/6
So, a solution for θ in the specified range is π/6.