What is the general solution for cos(θ)=sec(θ),0<= θ<2pi ?

Q: What is the general solution for cos(θ)=sec(θ),0<= θ<2pi ?

The general solution for cos(θ)=sec(θ),0<= θ<2pi is θ=0,θ=pi

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cos(θ)=sec(θ),0<= θ<2pi

Solution

cos (θ) = sec (θ), 0 \leq θ < 2 π

Solution

θ = 0, θ = π

Degrees

θ = 0^{\circ}, θ = 18 0^{\circ}

Solution steps

cos (θ) = sec (θ), 0 \leq θ < 2 π

Subtract

sec (θ)

from both sides

cos (θ) - sec (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

Subtract

1

from both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 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, 0 \leq θ < 2 π : θ = 0

sec (θ) = 1, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = 0

sec (θ) = - 1, 0 \leq θ < 2 π : θ = π

sec (θ) = - 1, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = π

Combine all the solutions

θ = 0, θ = π

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sqrt(2)sin(3x)=1,0<= x<= 4pi

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\frac{4}{cos ( x )} - 6 = tan (x)

tan(2a)cot(a+20)=1

tan (2 a) cot (a + 2 0^{\circ}) = 1

Frequently Asked Questions (FAQ)

What is the general solution for cos(θ)=sec(θ),0<= θ<2pi ?
The general solution for cos(θ)=sec(θ),0<= θ<2pi is θ=0,θ=pi