If 3 sec²θ-5tan θ-4=0 find the general solution to this equation​

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

здрасти

step-by-step explanation:

the answer is a

step-by-step explanation:

because your just changing the position of where the shape is