What is the general solution for tan^2(θ)+5tan(θ)+6=0 ?

The general solution for tan^2(θ)+5tan(θ)+6=0 is θ=-1.10714…+pin,θ=-1.24904…+pin

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Popular >

tan^2(θ)+5tan(θ)+6=0

Solution

tan^{2} (θ) + 5 tan (θ) + 6 = 0

Solution

θ = - 1.10714 \dots + πn, θ = - 1.24904 \dots + πn

Degrees

θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n, θ = - 71.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 5 tan (θ) + 6 = 0

Solve by substitution

tan^{2} (θ) + 5 tan (θ) + 6 = 0

Let:

tan (θ) = u

u^{2} + 5 u + 6 = 0

u^{2} + 5 u + 6 = 0 : u = - 2, u = - 3

u^{2} + 5 u + 6 = 0

Solve with the quadratic formula

u^{2} + 5 u + 6 = 0

Quadratic Equation Formula:

For

a = 1, b = 5, c = 6

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 6 = 1

5^{2} - 4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

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

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

Subtract the numbers:

- 5 - 1 = - 6

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6}{2}

Apply the fraction rule:

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

= - \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3

The solutions to the quadratic equation are:

u = - 2, u = - 3

Substitute back

u = tan (θ)

tan (θ) = - 2, tan (θ) = - 3

tan (θ) = - 2, tan (θ) = - 3

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) = - 3 : θ = arctan (- 3) + πn

tan (θ) = - 3

Apply trig inverse properties

tan (θ) = - 3

General solutions for

tan (θ) = - 3

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

θ = arctan (- 3) + πn

θ = arctan (- 3) + πn

Combine all the solutions

θ = arctan (- 2) + πn, θ = arctan (- 3) + πn

Show solutions in decimal form

θ = - 1.10714 \dots + πn, θ = - 1.24904 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+5tan(θ)+6=0 ?
The general solution for tan^2(θ)+5tan(θ)+6=0 is θ=-1.10714…+pin,θ=-1.24904…+pin