Find the solution set of each equation if 0° ≤ θ < 360°. (Please explain it with proper solutions)

Question

3(cot θ)^2 + 2 cot θ − 4 = 0

Answer 1

3q²+2q-4=0

q=either (-1/3)+/-(1/6)√(4+48)

aka cot(x)=either (-1/3)+/-(1/6)√(4+48)

x=either

cot^-1( (-1/3)+(1/6)√(4+48)+n*pi ,,,,n=0,+/-1,...

or cot^-1( (-1/3)-(1/6)√(4+48)+n*pi n=0,+/-1,...

to keep within range use proper n combination with x choices

Answer 2

3 cot² θ + 2 cot θ - 4 = 0

(3 cot θ - 2)(cot θ + 2) = 0

Use Zero Product Rule (Algebra 2) to solve for θ.