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

The general solution for sec^2(θ)=1 is θ=2pin,θ=pi+2pin

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sec^2(θ)=1

sec^{2} (θ) = 1

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sec^{2} (θ) = 1

Solve by substitution

sec^{2} (θ) = 1

Let:

sec (θ) = u

u^{2} = 1

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = sec (θ)

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1 : θ = 2 πn

sec (θ) = 1

General solutions for

sec (θ) = 1

sec (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

sec (θ) = - 1 : θ = π + 2 πn

sec (θ) = - 1

General solutions for

sec (θ) = - 1

sec (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

cos (\frac{x + π}{2}) = sin (x)

2 sin (2 x) - 1 = 0, 0 \leq x < 2 π

cos (A) = - \frac{4}{5}

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

sin (3 x) cos (x) - cos (3 x) sin (x) = 0

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