What is the general solution for tan^2(θ)+4sec(θ)=-5 ?

The general solution for tan^2(θ)+4sec(θ)=-5 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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tan^2(θ)+4sec(θ)=-5

Solution

tan^{2} (θ) + 4 sec (θ) = - 5

Solution

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

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

tan^{2} (θ) + 4 sec (θ) = - 5

Subtract

- 5

from both sides

tan^{2} (θ) + 4 sec (θ) + 5 = 0

Rewrite using trig identities

5 + tan^{2} (θ) + 4 sec (θ)

Use the Pythagorean identity:

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

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

= 5 + sec^{2} (θ) - 1 + 4 sec (θ)

Simplify

5 + sec^{2} (θ) - 1 + 4 sec (θ) : sec^{2} (θ) + 4 sec (θ) + 4

5 + sec^{2} (θ) - 1 + 4 sec (θ)

Group like terms

= sec^{2} (θ) + 4 sec (θ) + 5 - 1

Add/Subtract the numbers:

5 - 1 = 4

= sec^{2} (θ) + 4 sec (θ) + 4

= sec^{2} (θ) + 4 sec (θ) + 4

4 + sec^{2} (θ) + 4 sec (θ) = 0

Solve by substitution

4 + sec^{2} (θ) + 4 sec (θ) = 0

Let:

sec (θ) = u

4 + u^{2} + 4 u = 0

4 + u^{2} + 4 u = 0 : u = - 2

4 + u^{2} + 4 u = 0

Write in the standard form

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

u^{2} + 4 u + 4 = 0

Solve with the quadratic formula

u^{2} + 4 u + 4 = 0

Quadratic Equation Formula:

For

a = 1, b = 4, c = 4

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

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

4^{2} - 4 \cdot 1 \cdot 4 = 0

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

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 1}

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

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

\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

u = - 2

The solution to the quadratic equation is:

u = - 2

Substitute back

u = sec (θ)

sec (θ) = - 2

sec (θ) = - 2

sec (θ) = - 2 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

sec (θ) = - 2

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

Combine all the solutions

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

Popular Examples

0=tan(θ)

0 = tan (θ)

cos(θ)= 4/9

cos (θ) = \frac{4}{9}

-6cos(x)=0

- 6 cos (x) = 0

tan(x)3cot(x)=5sec(x)

tan (x) 3 cot (x) = 5 sec (x)

10cos(θ)=5

10 cos (θ) = 5

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+4sec(θ)=-5 ?
The general solution for tan^2(θ)+4sec(θ)=-5 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin