How would you solve this trig question (unit circle)

Solve trig equation for [0,360]
Tan^2 theta = 3

and

Solve trig equation for [0, 4pi]
Tan theta = - sqrt(3)

Please, if you could do it step by step, I keep having trouble with tan!
Thank you very much in advance!!

1) Solve trig equation for [0,360]
Tan^2 theta = 3

Tan(theta) = +/- sqrt(3)

The tangent = sqrt(3) at all 60 degree angles (from your unit circle), since sin(60) = sqrt(3)/2 and cos(60)= 1/2, and tan= sin/cos.

So theta = 60, 120, 240, or 300 degrees


and

2) Solve trig equation for [0, 4pi]
Tan theta = - sqrt(3)

Tanget is sqr(3) at 60 degree angles, and is negative in quadrants II And IV.

So 2pi/3 and 5pi/3
Then to go around again to 4pi: add 2pi, or 6pi/3:
8pi/3 and 11pi/3

Hoping this helps!

1.) tan^2 θ = 3

=> tan θ = ± √3
Therefore general solution: θ = nπ ± k,
where k is particular solution for (tan θ = √3)
Thus, general solution is: θ = nπ ± π/3
Substitute various values of 'n' to get solution in desired interval i.e [0,2π]

2.) tan θ = -√3

General solution is given by: θ = nπ + t
where 't' is particular solution for (tan θ = - √3)
Thus, general solution is: θ = nπ + (-π/3) = nπ - (π/3)
Substitute 'n' to obtain desired solution in [0,4π]

1) tan ^ 2 theta = 3
tan(theta) = sqrt(3)
theta = 60, 240
2) theta = tan^(-1)(-sqrt(3))
theta = 2pi/3, 5pi/3