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

The general solution for tan^2(θ)-2sec(θ)=2 is θ=1.23095…+2pin,θ=2pi-1.23095…+2pin,θ=pi+2pin

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tan^2(θ)-2sec(θ)=2

Solution

tan^{2} (θ) - 2 sec (θ) = 2

Solution

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn, θ = π + 2 πn

Degrees

θ = 70.52877 \dots^{\circ} + 36 0^{\circ} n, θ = 289.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

tan^{2} (θ) - 2 sec (θ) = 2

Subtract

2

from both sides

tan^{2} (θ) - 2 sec (θ) - 2 = 0

Rewrite using trig identities

- 2 + tan^{2} (θ) - 2 sec (θ)

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= - 2 + sec^{2} (θ) - 1 - 2 sec (θ)

Simplify

- 2 + sec^{2} (θ) - 1 - 2 sec (θ) : sec^{2} (θ) - 2 sec (θ) - 3

- 2 + sec^{2} (θ) - 1 - 2 sec (θ)

Group like terms

= sec^{2} (θ) - 2 sec (θ) - 2 - 1

Subtract the numbers:

- 2 - 1 = - 3

= sec^{2} (θ) - 2 sec (θ) - 3

= sec^{2} (θ) - 2 sec (θ) - 3

- 3 + sec^{2} (θ) - 2 sec (θ) = 0

Solve by substitution

- 3 + sec^{2} (θ) - 2 sec (θ) = 0

Let:

sec (θ) = u

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

- 3 + u^{2} - 2 u = 0 : u = 3, u = - 1

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

Write in the standard form

a x^{2} + b x + c = 0

u^{2} - 2 u - 3 = 0

Solve with the quadratic formula

u^{2} - 2 u - 3 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

u = \frac{- ( - 2 ) + 4}{2 \cdot 1} : 3

\frac{- ( - 2 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

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

Add the numbers:

2 + 4 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

u = \frac{- ( - 2 ) - 4}{2 \cdot 1} : - 1

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 4 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 3, u = - 1

Substitute back

u = sec (θ)

sec (θ) = 3, sec (θ) = - 1

sec (θ) = 3, sec (θ) = - 1

sec (θ) = 3 : θ = arcsec (3) + 2 πn, θ = 2 π - arcsec (3) + 2 πn

sec (θ) = 3

Apply trig inverse properties

sec (θ) = 3

General solutions for

sec (θ) = 3

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

θ = arcsec (3) + 2 πn, θ = 2 π - arcsec (3) + 2 πn

θ = arcsec (3) + 2 πn, θ = 2 π - arcsec (3) + 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 = arcsec (3) + 2 πn, θ = 2 π - arcsec (3) + 2 πn, θ = π + 2 πn

Show solutions in decimal form

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn, θ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)-2sec(θ)=2 ?
The general solution for tan^2(θ)-2sec(θ)=2 is θ=1.23095…+2pin,θ=2pi-1.23095…+2pin,θ=pi+2pin