What is the general solution for 6tan^2(θ)+11tan(θ)=4tan(θ)-2 ?

The general solution for 6tan^2(θ)+11tan(θ)=4tan(θ)-2 is θ=-0.46364…+pin,θ=-0.58800…+pin

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Popular >

6tan^2(θ)+11tan(θ)=4tan(θ)-2

Solution

6 tan^{2} (θ) + 11 tan (θ) = 4 tan (θ) - 2

Solution

θ = - 0.46364 \dots + πn, θ = - 0.58800 \dots + πn

Degrees

θ = - 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 33.69006 \dots^{\circ} + 18 0^{\circ} n

Solution steps

6 tan^{2} (θ) + 11 tan (θ) = 4 tan (θ) - 2

Solve by substitution

6 tan^{2} (θ) + 11 tan (θ) = 4 tan (θ) - 2

Let:

tan (θ) = u

6 u^{2} + 11 u = 4 u - 2

6 u^{2} + 11 u = 4 u - 2 : u = - \frac{1}{2}, u = - \frac{2}{3}

6 u^{2} + 11 u = 4 u - 2

Move

2

to the left side

6 u^{2} + 11 u = 4 u - 2

Add

2

to both sides

6 u^{2} + 11 u + 2 = 4 u - 2 + 2

Simplify

6 u^{2} + 11 u + 2 = 4 u

6 u^{2} + 11 u + 2 = 4 u

Move

4 u

to the left side

6 u^{2} + 11 u + 2 = 4 u

Subtract

4 u

from both sides

6 u^{2} + 11 u + 2 - 4 u = 4 u - 4 u

Simplify

6 u^{2} + 7 u + 2 = 0

6 u^{2} + 7 u + 2 = 0

Solve with the quadratic formula

6 u^{2} + 7 u + 2 = 0

Quadratic Equation Formula:

For

a = 6, b = 7, c = 2

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 6 \cdot 2}{2 \cdot 6}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 6 \cdot 2}{2 \cdot 6}

7^{2} - 4 \cdot 6 \cdot 2 = 1

7^{2} - 4 \cdot 6 \cdot 2

Multiply the numbers:

4 \cdot 6 \cdot 2 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

Subtract the numbers:

49 - 48 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 7 \pm 1}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 7 + 1}{2 \cdot 6}, u_{2} = \frac{- 7 - 1}{2 \cdot 6}

u = \frac{- 7 + 1}{2 \cdot 6} : - \frac{1}{2}

\frac{- 7 + 1}{2 \cdot 6}

Add/Subtract the numbers:

- 7 + 1 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

u = \frac{- 7 - 1}{2 \cdot 6} : - \frac{2}{3}

\frac{- 7 - 1}{2 \cdot 6}

Subtract the numbers:

- 7 - 1 = - 8

= \frac{- 8}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 8}{12}

Apply the fraction rule:

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

= - \frac{8}{12}

Cancel the common factor:

4

= - \frac{2}{3}

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = - \frac{2}{3}

Substitute back

u = tan (θ)

tan (θ) = - \frac{1}{2}, tan (θ) = - \frac{2}{3}

tan (θ) = - \frac{1}{2}, tan (θ) = - \frac{2}{3}

tan (θ) = - \frac{1}{2} : θ = arctan (- \frac{1}{2}) + πn

tan (θ) = - \frac{1}{2}

Apply trig inverse properties

tan (θ) = - \frac{1}{2}

General solutions for

tan (θ) = - \frac{1}{2}

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

θ = arctan (- \frac{1}{2}) + πn

θ = arctan (- \frac{1}{2}) + πn

tan (θ) = - \frac{2}{3} : θ = arctan (- \frac{2}{3}) + πn

tan (θ) = - \frac{2}{3}

Apply trig inverse properties

tan (θ) = - \frac{2}{3}

General solutions for

tan (θ) = - \frac{2}{3}

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

θ = arctan (- \frac{2}{3}) + πn

θ = arctan (- \frac{2}{3}) + πn

Combine all the solutions

θ = arctan (- \frac{1}{2}) + πn, θ = arctan (- \frac{2}{3}) + πn

Show solutions in decimal form

θ = - 0.46364 \dots + πn, θ = - 0.58800 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 6tan^2(θ)+11tan(θ)=4tan(θ)-2 ?
The general solution for 6tan^2(θ)+11tan(θ)=4tan(θ)-2 is θ=-0.46364…+pin,θ=-0.58800…+pin