What is the general solution for 4tan^2(θ)+tan(θ)=-4tan(θ)-1 ?

The general solution for 4tan^2(θ)+tan(θ)=-4tan(θ)-1 is θ=-0.24497…+pin,θ=(3pi)/4+pin

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4tan^2(θ)+tan(θ)=-4tan(θ)-1

Solution

4 tan^{2} (θ) + tan (θ) = - 4 tan (θ) - 1

Solution

θ = - 0.24497 \dots + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = - 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

4 tan^{2} (θ) + tan (θ) = - 4 tan (θ) - 1

Solve by substitution

4 tan^{2} (θ) + tan (θ) = - 4 tan (θ) - 1

Let:

tan (θ) = u

4 u^{2} + u = - 4 u - 1

4 u^{2} + u = - 4 u - 1 : u = - \frac{1}{4}, u = - 1

4 u^{2} + u = - 4 u - 1

Move

1

to the left side

4 u^{2} + u = - 4 u - 1

Add

1

to both sides

4 u^{2} + u + 1 = - 4 u - 1 + 1

Simplify

4 u^{2} + u + 1 = - 4 u

4 u^{2} + u + 1 = - 4 u

Move

4 u

to the left side

4 u^{2} + u + 1 = - 4 u

Add

4 u

to both sides

4 u^{2} + u + 1 + 4 u = - 4 u + 4 u

Simplify

4 u^{2} + 5 u + 1 = 0

4 u^{2} + 5 u + 1 = 0

Solve with the quadratic formula

4 u^{2} + 5 u + 1 = 0

Quadratic Equation Formula:

For

a = 4, b = 5, c = 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

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

5^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 5 + 3}{2 \cdot 4}, u_{2} = \frac{- 5 - 3}{2 \cdot 4}

u = \frac{- 5 + 3}{2 \cdot 4} : - \frac{1}{4}

\frac{- 5 + 3}{2 \cdot 4}

Add/Subtract the numbers:

- 5 + 3 = - 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2}{8}

Apply the fraction rule:

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

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

u = \frac{- 5 - 3}{2 \cdot 4} : - 1

\frac{- 5 - 3}{2 \cdot 4}

Subtract the numbers:

- 5 - 3 = - 8

= \frac{- 8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8}{8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - \frac{1}{4}, u = - 1

Substitute back

u = tan (θ)

tan (θ) = - \frac{1}{4}, tan (θ) = - 1

tan (θ) = - \frac{1}{4}, tan (θ) = - 1

tan (θ) = - \frac{1}{4} : θ = arctan (- \frac{1}{4}) + πn

tan (θ) = - \frac{1}{4}

Apply trig inverse properties

tan (θ) = - \frac{1}{4}

General solutions for

tan (θ) = - \frac{1}{4}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- \frac{1}{4}) + πn

θ = arctan (- \frac{1}{4}) + πn

tan (θ) = - 1 : θ = \frac{3 π}{4} + πn

tan (θ) = - 1

General solutions for

tan (θ) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = arctan (- \frac{1}{4}) + πn, θ = \frac{3 π}{4} + πn

Show solutions in decimal form

θ = - 0.24497 \dots + πn, θ = \frac{3 π}{4} + πn

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7 tan^{2} (x) - 21 = 0

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Frequently Asked Questions (FAQ)

What is the general solution for 4tan^2(θ)+tan(θ)=-4tan(θ)-1 ?
The general solution for 4tan^2(θ)+tan(θ)=-4tan(θ)-1 is θ=-0.24497…+pin,θ=(3pi)/4+pin