Please help me solve these trigonometric functions?

Question

Solve the equation for exact solutions over the interval (0,360)

1. 4 tan θ + √3 = tan θ
2. cosθ = 1 + 3cos θ

Answer 1

1. 4 tan θ + √3 = tan θ subtract 4 tan θ from both sides
2. √3 = -3tan θ divide both sides by -3
3. tan θ= -√3/3
4. tan is negative in the II and IV Quadrants so
5. θ=150 and 330
6. _----__----------------------------------------
7. cosθ = 1 + 3cos θ subtract 3cos θ from both sides
8. -2 cos θ=1 divide both sides by -2
9. cos θ= -1/2 cosine is negative in Quadrant II and III
10. θ=120 and 240

Answer 2

Starting with equation 1 we can isolate the tan(θ) term as follows:

4 tan(θ) + sqrt(3) = tan(θ)

3 tan(θ) = -sqrt(3)

tan(θ) = -sqrt(3) / 3 = -1 / sqrt(3)

Therefore

θ = arctan(-1/sqrt(3)) = -30.0 degrees = 330 degrees.

Starting with equation 2 we isolate the cos(θ term as follows:

cos(θ) = 1 + 3 cos(θ)

2cos(θ) = -1

cos(θ) = -1/2

Therefore

θ = arccos(-1/2) = 120.0 degrees.

Answer 3