What is the general solution for tan^2(θ)+9tan(θ)+18=0 ?

The general solution for tan^2(θ)+9tan(θ)+18=0 is θ=-1.24904…+pin,θ=-1.40564…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(θ)+9tan(θ)+18=0

Solution

tan^{2} (θ) + 9 tan (θ) + 18 = 0

Solution

θ = - 1.24904 \dots + πn, θ = - 1.40564 \dots + πn

Degrees

θ = - 71.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 80.53767 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 9 tan (θ) + 18 = 0

Solve by substitution

tan^{2} (θ) + 9 tan (θ) + 18 = 0

Let:

tan (θ) = u

u^{2} + 9 u + 18 = 0

u^{2} + 9 u + 18 = 0 : u = - 3, u = - 6

u^{2} + 9 u + 18 = 0

Solve with the quadratic formula

u^{2} + 9 u + 18 = 0

Quadratic Equation Formula:

For

a = 1, b = 9, c = 18

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 18}{2 \cdot 1}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 18}{2 \cdot 1}

9^{2} - 4 \cdot 1 \cdot 18 = 3

9^{2} - 4 \cdot 1 \cdot 18

Multiply the numbers:

4 \cdot 1 \cdot 18 = 72

= 9^{2} - 72

9^{2} = 81

= 81 - 72

Subtract the numbers:

81 - 72 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 9 \pm 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 9 + 3}{2 \cdot 1}, u_{2} = \frac{- 9 - 3}{2 \cdot 1}

u = \frac{- 9 + 3}{2 \cdot 1} : - 3

\frac{- 9 + 3}{2 \cdot 1}

Add/Subtract the numbers:

- 9 + 3 = - 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

u = \frac{- 9 - 3}{2 \cdot 1} : - 6

\frac{- 9 - 3}{2 \cdot 1}

Subtract the numbers:

- 9 - 3 = - 12

= \frac{- 12}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 12}{2}

Apply the fraction rule:

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

= - \frac{12}{2}

Divide the numbers:

\frac{12}{2} = 6

= - 6

The solutions to the quadratic equation are:

u = - 3, u = - 6

Substitute back

u = tan (θ)

tan (θ) = - 3, tan (θ) = - 6

tan (θ) = - 3, tan (θ) = - 6

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

tan (θ) = - 6 : θ = arctan (- 6) + πn

tan (θ) = - 6

Apply trig inverse properties

tan (θ) = - 6

General solutions for

tan (θ) = - 6

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

θ = arctan (- 6) + πn

θ = arctan (- 6) + πn

Combine all the solutions

θ = arctan (- 3) + πn, θ = arctan (- 6) + πn

Show solutions in decimal form

θ = - 1.24904 \dots + πn, θ = - 1.40564 \dots + πn

Popular Examples

cos^2(θ)-sin^2(θ)=-sin(θ)+1

cos^{2} (θ) - sin^{2} (θ) = - sin (θ) + 1

2sin(θ)+6=7

2 sin (θ) + 6 = 7

8cos(12x)+4=-4

8 cos (12 x) + 4 = - 4

12tan^3(x)=4tan(x)

12 tan^{3} (x) = 4 tan (x)

cos(2θ)=3sin(θ)-1

cos (2 θ) = 3 sin (θ) - 1

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+9tan(θ)+18=0 ?
The general solution for tan^2(θ)+9tan(θ)+18=0 is θ=-1.24904…+pin,θ=-1.40564…+pin