What is the general solution for tan^2(θ)+tan(θ)=6,0<= θ<180 ?

Q: What is the general solution for tan^2(θ)+tan(θ)=6,0<= θ<180 ?

The general solution for tan^2(θ)+tan(θ)=6,0<= θ<180 is θ=1.10714…,θ=-1.24904…+180

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(θ)+tan(θ)=6,0<= θ<180

Solution

tan^{2} (θ) + tan (θ) = 6, 0 \leq θ < 18 0^{\circ}

Solution

θ = 1.10714 \dots, θ = - 1.24904 \dots + 18 0^{\circ}

Radians

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

Solution steps

tan^{2} (θ) + tan (θ) = 6, 0 \leq θ < 18 0^{\circ}

Solve by substitution

tan^{2} (θ) + tan (θ) = 6

Let:

tan (θ) = u

u^{2} + u = 6

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

u^{2} + u = 6

Move

6

to the left side

u^{2} + u = 6

Subtract

6

from both sides

u^{2} + u - 6 = 6 - 6

Simplify

u^{2} + u - 6 = 0

u^{2} + u - 6 = 0

Solve with the quadratic formula

u^{2} + u - 6 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 6

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 6)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 6)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 1 + 5 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

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

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

Subtract the numbers:

- 1 - 5 = - 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, 0 \leq θ < 18 0^{\circ} : θ = arctan (2)

tan (θ) = 2, 0 \leq θ < 18 0^{\circ}

Apply trig inverse properties

tan (θ) = 2

General solutions for

tan (θ) = 2

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

θ = arctan (2) + 18 0^{\circ} n

θ = arctan (2) + 18 0^{\circ} n

Solutions for the range

0 \leq θ < 18 0^{\circ}

θ = arctan (2)

tan (θ) = - 3, 0 \leq θ < 18 0^{\circ} : θ = - arctan (3) + 18 0^{\circ}

tan (θ) = - 3, 0 \leq θ < 18 0^{\circ}

Apply trig inverse properties

tan (θ) = - 3

General solutions for

tan (θ) = - 3

tan (x) = - a \Rightarrow x = arctan (- a) + 18 0^{\circ} n

θ = arctan (- 3) + 18 0^{\circ} n

θ = arctan (- 3) + 18 0^{\circ} n

Solutions for the range

0 \leq θ < 18 0^{\circ}

θ = - arctan (3) + 18 0^{\circ}

Combine all the solutions

θ = arctan (2), θ = - arctan (3) + 18 0^{\circ}

Show solutions in decimal form

θ = 1.10714 \dots, θ = - 1.24904 \dots + 18 0^{\circ}

Popular Examples

sin(2x)=5sin(x)

sin (2 x) = 5 sin (x)

2cos^2(θ)-4cos(θ)-1=-5cos(θ)

2 cos^{2} (θ) - 4 cos (θ) - 1 = - 5 cos (θ)

-2sin^2(x)-sin(x)+3=0

- 2 sin^{2} (x) - sin (x) + 3 = 0

tan(x)=-sqrt(3)+2

tan (x) = - 3 + 2

3tan(x)-cot(x)=0

3 tan (x) - cot (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+tan(θ)=6,0<= θ<180 ?
The general solution for tan^2(θ)+tan(θ)=6,0<= θ<180 is θ=1.10714…,θ=-1.24904…+180