What is the general solution for sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi ?

Q: What is the general solution for sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi ?

The general solution for sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi is θ=2pi,θ=(8pi)/3

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Popular Trigonometry >

sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi

Solution

sec^{2} (θ) + sec (θ) = 2, 2 π \leq θ \leq 3 π

Solution

θ = 2 π, θ = \frac{8 π}{3}

Degrees

θ = 36 0^{\circ}, θ = 48 0^{\circ}

Solution steps

sec^{2} (θ) + sec (θ) = 2, 2 π \leq θ \leq 3 π

Solve by substitution

sec^{2} (θ) + sec (θ) = 2

Let:

sec (θ) = u

u^{2} + u = 2

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

u^{2} + u = 2

Move

2

to the left side

u^{2} + u = 2

Subtract

2

from both sides

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

Simplify

u^{2} + u - 2 = 0

u^{2} + u - 2 = 0

Solve with the quadratic formula

u^{2} + u - 2 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

1^{2} - 4 \cdot 1 \cdot (- 2)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 \cdot 1}, u_{2} = \frac{- 1 - 3}{2 \cdot 1}

u = \frac{- 1 + 3}{2 \cdot 1} : 1

\frac{- 1 + 3}{2 \cdot 1}

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 1 - 3}{2 \cdot 1} : - 2

\frac{- 1 - 3}{2 \cdot 1}

Subtract the numbers:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = 1, u = - 2

Substitute back

u = sec (θ)

sec (θ) = 1, sec (θ) = - 2

sec (θ) = 1, sec (θ) = - 2

sec (θ) = 1, 2 π \leq θ \leq 3 π : θ = 2 π

sec (θ) = 1, 2 π \leq θ \leq 3 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

2 π \leq θ \leq 3 π

θ = 2 π

sec (θ) = - 2, 2 π \leq θ \leq 3 π : θ = \frac{8 π}{3}

sec (θ) = - 2, 2 π \leq θ \leq 3 π

General solutions for

sec (θ) = - 2

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

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

Solutions for the range

2 π \leq θ \leq 3 π

θ = \frac{8 π}{3}

Combine all the solutions

θ = 2 π, θ = \frac{8 π}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi ?
The general solution for sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi is θ=2pi,θ=(8pi)/3