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

The general solution for tan^2(θ)+8tan(θ)+6=0 is θ=-0.69732…+pin,θ=-1.43207…+pin

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Popular >

tan^2(θ)+8tan(θ)+6=0

Solution

tan^{2} (θ) + 8 tan (θ) + 6 = 0

Solution

θ = - 0.69732 \dots + πn, θ = - 1.43207 \dots + πn

Degrees

θ = - 39.95366 \dots^{\circ} + 18 0^{\circ} n, θ = - 82.05172 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 8 tan (θ) + 6 = 0

Solve by substitution

tan^{2} (θ) + 8 tan (θ) + 6 = 0

Let:

tan (θ) = u

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

u^{2} + 8 u + 6 = 0 : u = - 4 + 10, u = - 4 - 10

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 8, c = 6

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

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

8^{2} - 4 \cdot 1 \cdot 6 = 210

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

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 8^{2} - 24

8^{2} = 64

= 64 - 24

Subtract the numbers:

64 - 24 = 40

= 40

Prime factorization of

40 : 2^{3} \cdot 5

40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 5

Apply radical rule:

= 2^{2} 2 \cdot 5

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 5

Refine

= 210

u_{1, 2} = \frac{- 8 \pm 2 10}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 8 + 2 10}{2 \cdot 1}, u_{2} = \frac{- 8 - 2 10}{2 \cdot 1}

u = \frac{- 8 + 2 10}{2 \cdot 1} : - 4 + 10

\frac{- 8 + 2 10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 8 + 2 10}{2}

Factor

- 8 + 210 : 2 (- 4 + 10)

- 8 + 210

Rewrite as

= - 2 \cdot 4 + 210

Factor out common term

2

= 2 (- 4 + 10)

= \frac{2 ( - 4 + 10 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 4 + 10

u = \frac{- 8 - 2 10}{2 \cdot 1} : - 4 - 10

\frac{- 8 - 2 10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 8 - 2 10}{2}

Factor

- 8 - 210 : - 2 (4 + 10)

- 8 - 210

Rewrite as

= - 2 \cdot 4 - 210

Factor out common term

2

= - 2 (4 + 10)

= - \frac{2 ( 4 + 10 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (4 + 10)

Negate

- (4 + 10) = - 4 - 10

= - 4 - 10

The solutions to the quadratic equation are:

u = - 4 + 10, u = - 4 - 10

Substitute back

u = tan (θ)

tan (θ) = - 4 + 10, tan (θ) = - 4 - 10

tan (θ) = - 4 + 10, tan (θ) = - 4 - 10

tan (θ) = - 4 + 10 : θ = arctan (- 4 + 10) + πn

tan (θ) = - 4 + 10

Apply trig inverse properties

tan (θ) = - 4 + 10

General solutions for

tan (θ) = - 4 + 10

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

θ = arctan (- 4 + 10) + πn

θ = arctan (- 4 + 10) + πn

tan (θ) = - 4 - 10 : θ = arctan (- 4 - 10) + πn

tan (θ) = - 4 - 10

Apply trig inverse properties

tan (θ) = - 4 - 10

General solutions for

tan (θ) = - 4 - 10

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

θ = arctan (- 4 - 10) + πn

θ = arctan (- 4 - 10) + πn

Combine all the solutions

θ = arctan (- 4 + 10) + πn, θ = arctan (- 4 - 10) + πn

Show solutions in decimal form

θ = - 0.69732 \dots + πn, θ = - 1.43207 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+8tan(θ)+6=0 ?
The general solution for tan^2(θ)+8tan(θ)+6=0 is θ=-0.69732…+pin,θ=-1.43207…+pin