solve for theta in the interval 0 to 360

Question

3 sec^2 theta - 5 tan theta= 1

Answer 1

3sec²θ - 5tanθ = 1

 

3(1 + tan²θ) - 5tanθ - 1 = 0

 

3tan²θ - 5tanθ + 2 = 0

 

Let u = tanθ     Then 3u² - 5u + 2 = 0

 

                                    (3u - 2)(u - 1) = 0

 

                                     u = 2/3 or u = 1

 

tanθ = 2/3   or tanθ = 1

 

If tanθ = 1, then θ = 45° or 225°

 

If tanθ = 2/3, then θ lies in quadrant 1 or quadrant 3

 

solution in quadrant 1:  θ = tan^-1(2/3) = 33.7°

 

solution in quadrant 3:  θ = 33.7° + 180° = 213.7°

 

Solutions:  45°, 225°, 33.7°, 213.7°