What is the general solution for 5sec(θ)-2sec^2(θ)=tan^2(θ)-1 ?

The general solution for 5sec(θ)-2sec^2(θ)=tan^2(θ)-1 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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5sec(θ)-2sec^2(θ)=tan^2(θ)-1

5 sec (θ) - 2 sec^{2} (θ) = tan^{2} (θ) - 1

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solution steps

5 sec (θ) - 2 sec^{2} (θ) = tan^{2} (θ) - 1

Subtract

tan^{2} (θ) - 1

from both sides

5 sec (θ) - 2 sec^{2} (θ) - tan^{2} (θ) + 1 = 0

Rewrite using trig identities

2 - 3 sec^{2} (θ) + 5 sec (θ) = 0

Solve by substitution

sec (θ) = - \frac{1}{3}, sec (θ) = 2

sec (θ) = - \frac{1}{3} :

No Solution

sec (θ) = 2 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (2 x) + cos (x) = 0, (0, 2 π)

cos (3 θ) = 1, 0 \leq θ \leq 2 π

3 \cdot sin (\frac{x}{2}) + cos (x) = 1

sin (2 x) = \frac{2}{3}

tan^{3} (θ) + tan^{2} (θ) - 3 tan (θ) - 3 = 0

What is the general solution for 5sec(θ)-2sec^2(θ)=tan^2(θ)-1 ?
The general solution for 5sec(θ)-2sec^2(θ)=tan^2(θ)-1 is θ= pi/3+2pin,θ=(5pi)/3+2pin