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

The general solution for 2tan^2(θ)+sec(θ)=1 is θ=2pin,θ=2.30052…+2pin,θ=-2.30052…+2pin

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2tan^2(θ)+sec(θ)=1

Solution

2 tan^{2} (θ) + sec (θ) = 1

Solution

θ = 2 πn, θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 131.81031 \dots^{\circ} + 36 0^{\circ} n, θ = - 131.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 tan^{2} (θ) + sec (θ) = 1

Subtract

1

from both sides

2 tan^{2} (θ) + sec (θ) - 1 = 0

Rewrite using trig identities

- 1 + sec (θ) + 2 tan^{2} (θ)

Use the Pythagorean identity:

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

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

= - 1 + sec (θ) + 2 (sec^{2} (θ) - 1)

Simplify

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

- 1 + sec (θ) + 2 (sec^{2} (θ) - 1)

Expand

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

2 (sec^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = sec^{2} (θ), c = 1

= 2 sec^{2} (θ) - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 sec^{2} (θ) - 2

= - 1 + sec (θ) + 2 sec^{2} (θ) - 2

Simplify

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

- 1 + sec (θ) + 2 sec^{2} (θ) - 2

Group like terms

= sec (θ) + 2 sec^{2} (θ) - 1 - 2

Subtract the numbers:

- 1 - 2 = - 3

= 2 sec^{2} (θ) + sec (θ) - 3

= 2 sec^{2} (θ) + sec (θ) - 3

= 2 sec^{2} (θ) + sec (θ) - 3

- 3 + sec (θ) + 2 sec^{2} (θ) = 0

Solve by substitution

- 3 + sec (θ) + 2 sec^{2} (θ) = 0

Let:

sec (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 \cdot 2 (- 3) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 3)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 2}, u_{2} = \frac{- 1 - 5}{2 \cdot 2}

u = \frac{- 1 + 5}{2 \cdot 2} : 1

\frac{- 1 + 5}{2 \cdot 2}

Add/Subtract the numbers:

- 1 + 5 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 1 - 5}{2 \cdot 2} : - \frac{3}{2}

\frac{- 1 - 5}{2 \cdot 2}

Subtract the numbers:

- 1 - 5 = - 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{4}

Apply the fraction rule:

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

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

The solutions to the quadratic equation are:

u = 1, u = - \frac{3}{2}

Substitute back

u = sec (θ)

sec (θ) = 1, sec (θ) = - \frac{3}{2}

sec (θ) = 1, sec (θ) = - \frac{3}{2}

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) = - \frac{3}{2} : θ = arcsec (- \frac{3}{2}) + 2 πn, θ = - arcsec (- \frac{3}{2}) + 2 πn

sec (θ) = - \frac{3}{2}

Apply trig inverse properties

sec (θ) = - \frac{3}{2}

General solutions for

sec (θ) = - \frac{3}{2}

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

θ = arcsec (- \frac{3}{2}) + 2 πn, θ = - arcsec (- \frac{3}{2}) + 2 πn

θ = arcsec (- \frac{3}{2}) + 2 πn, θ = - arcsec (- \frac{3}{2}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = arcsec (- \frac{3}{2}) + 2 πn, θ = - arcsec (- \frac{3}{2}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = 2.30052 \dots + 2 πn, θ = - 2.30052 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(θ)+sec(θ)=1 ?
The general solution for 2tan^2(θ)+sec(θ)=1 is θ=2pin,θ=2.30052…+2pin,θ=-2.30052…+2pin