Question 1026572
(tan θ − 3)(9 sin^2 θ − 1) = 0

When we have A * B = 0, the roots are A=0, B=0 so we start with 

Tan θ - 3 = 0

Tan θ = 3 

θ = 71.565 degrees +/- 180N (tangen has a 180 period

(9 sin^2 θ − 1) = 0

This factors to

(3 sin θ − 1)(3 sin θ + 1) = 0 and 

3 sin θ − 1 = 0

3 sin θ = 1

sin θ = 1/3

θ = 19.4712 degrees, and 160.5288 degrees (+/-360N on both answers)

3 sin θ + 1 = 0

3 sin θ = -1

sin θ = -1/3

θ = -19.4712 degrees, and -160.5288 degrees (+/-360N on both answers)