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=1.24904…,θ=-1.10714…+180

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tan^2(θ)-tan(θ)=6,0<= θ<180

Solution

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

Solution

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

Radians

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

Solution steps

tan^{2} (θ) - tan (θ) = 6, 0^{\circ} \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 = 3, u = - 2

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

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 6 = 24

4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 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} : 3

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 5 = 6

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 5 = - 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

The solutions to the quadratic equation are:

u = 3, u = - 2

Substitute back

u = tan (θ)

tan (θ) = 3, tan (θ) = - 2

tan (θ) = 3, tan (θ) = - 2

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

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)

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

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) + 18 0^{\circ}

Combine all the solutions

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

Show solutions in decimal form

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

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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=1.24904…,θ=-1.10714…+180