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

The general solution for -tan^2(θ)=-1+sec(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=2pin

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-tan^2(θ)=-1+sec(θ)

- tan^{2} (θ) = - 1 + sec (θ)

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = 2 πn

Solution steps

- tan^{2} (θ) = - 1 + sec (θ)

Subtract

- 1 + sec (θ)

from both sides

- tan^{2} (θ) + 1 - sec (θ) = 0

Rewrite using trig identities

2 - sec (θ) - sec^{2} (θ) = 0

Solve by substitution

sec (θ) = - 2, sec (θ) = 1

sec (θ) = - 2 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

sec (θ) = 1 : θ = 2 πn

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = 2 πn

x, y = sin (x^{c o s (y)})

3 sec (θ) + 3 = 0

12 \cdot 9.8 \cdot sin (a) - 0.6 \cdot 12 \cdot 9.8 \cdot cos (a) = 12 \cdot 1.79

sin (x) (sin (x) - 1) = 5 cos^{2} (x) - 4

v, a = arctan (\frac{v ^{2}}{g R})

What is the general solution for -tan^2(θ)=-1+sec(θ) ?
The general solution for -tan^2(θ)=-1+sec(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=2pin