What is the general solution for tan^2(θ)+6tan(θ)+8=0 ?

The general solution for tan^2(θ)+6tan(θ)+8=0 is θ=-1.10714…+pin,θ=-1.32581…+pin

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Popular >

tan^2(θ)+6tan(θ)+8=0

Solution

tan^{2} (θ) + 6 tan (θ) + 8 = 0

Solution

θ = - 1.10714 \dots + πn, θ = - 1.32581 \dots + πn

Degrees

θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n, θ = - 75.96375 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 6 tan (θ) + 8 = 0

Solve by substitution

tan^{2} (θ) + 6 tan (θ) + 8 = 0

Let:

tan (θ) = u

u^{2} + 6 u + 8 = 0

u^{2} + 6 u + 8 = 0 : u = - 2, u = - 4

u^{2} + 6 u + 8 = 0

Solve with the quadratic formula

u^{2} + 6 u + 8 = 0

Quadratic Equation Formula:

For

a = 1, b = 6, c = 8

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot 8 = 2

6^{2} - 4 \cdot 1 \cdot 8

Multiply the numbers:

4 \cdot 1 \cdot 8 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

Subtract the numbers:

36 - 32 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- 6 \pm 2}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 6 + 2}{2 \cdot 1}, u_{2} = \frac{- 6 - 2}{2 \cdot 1}

u = \frac{- 6 + 2}{2 \cdot 1} : - 2

\frac{- 6 + 2}{2 \cdot 1}

Add/Subtract the numbers:

- 6 + 2 = - 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

u = \frac{- 6 - 2}{2 \cdot 1} : - 4

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

Subtract the numbers:

- 6 - 2 = - 8

= \frac{- 8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 8}{2}

Apply the fraction rule:

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

= - \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= - 4

The solutions to the quadratic equation are:

u = - 2, u = - 4

Substitute back

u = tan (θ)

tan (θ) = - 2, tan (θ) = - 4

tan (θ) = - 2, tan (θ) = - 4

tan (θ) = - 2 : θ = arctan (- 2) + πn

tan (θ) = - 2

Apply trig inverse properties

tan (θ) = - 2

General solutions for

tan (θ) = - 2

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

θ = arctan (- 2) + πn

θ = arctan (- 2) + πn

tan (θ) = - 4 : θ = arctan (- 4) + πn

tan (θ) = - 4

Apply trig inverse properties

tan (θ) = - 4

General solutions for

tan (θ) = - 4

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

θ = arctan (- 4) + πn

θ = arctan (- 4) + πn

Combine all the solutions

θ = arctan (- 2) + πn, θ = arctan (- 4) + πn

Show solutions in decimal form

θ = - 1.10714 \dots + πn, θ = - 1.32581 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+6tan(θ)+8=0 ?
The general solution for tan^2(θ)+6tan(θ)+8=0 is θ=-1.10714…+pin,θ=-1.32581…+pin