How do I solve cot2θ+3cotθ=0?

1 Answer
Jun 23, 2018

θ=±π2+2kπ,θ1=π6+2kπ,θ2=5π6+2kπ

Explanation:

Rewriting your equation in the form
cot(θ)(cot(θ)+3)=0
so we get two cases:

cot(θ)=0
since

cot(x)=cos(x)sin(x)=0if cos(x)=0

we get

θ=±π2+2kπ
the other case:
cot(θ)=3
θ1=π6+2kπ
θ2=5π6+2kπ