3 sec²(θ) - 5 tan(θ) - 4 = 0
Recall the Pythagorean identity,
cos²(θ) + sin²(θ) = 1.
Multiplying both sides by 1/cos²(θ) gives another form of the identity,
1 + tan²(θ) = sec²(θ).
Then the equation becomes quadratic in tan(θ):
3 (1 + tan²(θ)) - 5 tan(θ) - 4 = 0
3 tan²(θ) - 5 tan(θ) - 1 = 0
I'll solve by completing the square.
tan²(θ) - 5/3 tan(θ)) - 1/3 = 0
tan²(θ) - 5/3 tan(θ) = 1/3
tan²(θ) - 5/3 tan(θ) + 25/36 = 1/3 + 25/36
(tan(θ) - 5/6)² = 37/36
tan(θ) - 5/6 = ±√37/6
tan(θ) = (5 ± √37)/6
Take the inverse tangent of both sides:
θ = arctan((5 + √37)/6) + nπ or θ = arctan((5 - √37)/6) + nπ
where n is any integer
