What is the general solution for sec(θ)=cos(θ) ?

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

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sec(θ)=cos(θ)

Solution

sec (θ) = cos (θ)

Solution

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 (θ) = cos (θ)

Subtract

cos (θ)

from both sides

sec (θ) - cos (θ) = 0

Rewrite using trig identities

- cos (θ) + sec (θ)

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

= - \frac{1}{sec ( θ )} + sec (θ)

- \frac{1}{sec ( θ )} + sec (θ) = 0

Solve by substitution

- \frac{1}{sec ( θ )} + sec (θ) = 0

Let:

sec (θ) = u

- \frac{1}{u} + u = 0

- \frac{1}{u} + u = 0 : u = 1, u = - 1

- \frac{1}{u} + u = 0

Multiply both sides by

u

- \frac{1}{u} + u = 0

Multiply both sides by

u

- \frac{1}{u} u + uu = 0 \cdot u

Simplify

- \frac{1}{u} u + uu = 0 \cdot u

Simplify

- \frac{1}{u} u : - 1

- \frac{1}{u} u

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot u}{u}

Cancel the common factor:

u

= - 1

Simplify

uu : u^{2}

uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

- 1 + u^{2} = 0

- 1 + u^{2} = 0

- 1 + u^{2} = 0

Solve

- 1 + u^{2} = 0 : u = 1, u = - 1

- 1 + u^{2} = 0

Move

1

to the right side

- 1 + u^{2} = 0

Add

1

to both sides

- 1 + u^{2} + 1 = 0 + 1

Simplify

u^{2} = 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

u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- \frac{1}{u} + u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

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Frequently Asked Questions (FAQ)

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