Need help with equations involving trigonometric functions!!

Question

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

1.) 3 sin θ = sin θ - √3

2.) 5cos θ = 3cos θ - 1

Answer 1

3 sin θ = sin θ - √3

subtract sin θ from both sides

2sin θ= -√3

Divide both sides by 2

sin θ= -√3/2

sin is negative in Quadrant III and IV

θ=240 and 300

5cos θ = 3cos θ - 1

Subtract 3cos θ from both sides

2cos θ = -1

Divide both sides by 2

cos θ = -1/2

cos is negative in Quadrant II and III

θ=120 and 240

Answer 2

Isolate sin(θ) in the first equation:

2 sin(θ) = -sqrt(3)

sin(θ) = -sqrt(3) / 2

Therefore

θ = arcsin(-sqrt(3)/2) = -60.0 degrees = 300 degrees.

Isolate cos(θ) in the second equation:

2 cos(θ) = -1

cos(θ) = -1/2

Therefore

θ = acrcos(-1/2) = 120.0 degrees