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

The general solution for -6tan^2(θ)+11tan(θ)-6=-2tan(θ) is θ=0.58800…+pin,θ=0.98279…+pin

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Popular >

-6tan^2(θ)+11tan(θ)-6=-2tan(θ)

Solution

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

Solution

θ = 0.58800 \dots + πn, θ = 0.98279 \dots + πn

Degrees

θ = 33.69006 \dots^{\circ} + 18 0^{\circ} n, θ = 56.30993 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (θ) = u

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

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

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

Move

2 u

to the left side

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

Add

2 u

to both sides

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

Simplify

- 6 u^{2} + 13 u - 6 = 0

- 6 u^{2} + 13 u - 6 = 0

Solve with the quadratic formula

- 6 u^{2} + 13 u - 6 = 0

Quadratic Equation Formula:

For

a = - 6, b = 13, c = - 6

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 ( - 6 ) ( - 6 )}{2 ( - 6 )}

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 ( - 6 ) ( - 6 )}{2 ( - 6 )}

1 3^{2} - 4 (- 6) (- 6) = 5

1 3^{2} - 4 (- 6) (- 6)

Apply rule

- (- a) = a

= 1 3^{2} - 4 \cdot 6 \cdot 6

Multiply the numbers:

4 \cdot 6 \cdot 6 = 144

= 1 3^{2} - 144

1 3^{2} = 169

= 169 - 144

Subtract the numbers:

169 - 144 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 13 \pm 5}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- 13 + 5}{2 ( - 6 )}, u_{2} = \frac{- 13 - 5}{2 ( - 6 )}

u = \frac{- 13 + 5}{2 ( - 6 )} : \frac{2}{3}

\frac{- 13 + 5}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 13 + 5}{- 2 \cdot 6}

Add/Subtract the numbers:

- 13 + 5 = - 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}

u = \frac{- 13 - 5}{2 ( - 6 )} : \frac{3}{2}

\frac{- 13 - 5}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 13 - 5}{- 2 \cdot 6}

Subtract the numbers:

- 13 - 5 = - 18

= \frac{- 18}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 18}{- 12}

Apply the fraction rule:

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

= \frac{18}{12}

Cancel the common factor:

6

= \frac{3}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = tan (θ)

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

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

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

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

tan (θ) = \frac{3}{2}

Apply trig inverse properties

tan (θ) = \frac{3}{2}

General solutions for

tan (θ) = \frac{3}{2}

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

θ = arctan (\frac{3}{2}) + πn

θ = arctan (\frac{3}{2}) + πn

Combine all the solutions

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

Show solutions in decimal form

θ = 0.58800 \dots + πn, θ = 0.98279 \dots + πn

Popular Examples

(sin(x)+1)^2+cos^2(x)=2

(sin (x) + 1)^{2} + cos^{2} (x) = 2

12sin^2(x)-cos(x)-6=0

12 sin^{2} (x) - cos (x) - 6 = 0

tan(x/2)-1=0

tan (\frac{x}{2}) - 1 = 0

sin(θ)=-cos(θ)

sin (θ) = - cos (θ)

cot^2(θ)+csc(θ)=1

cot^{2} (θ) + csc (θ) = 1

Frequently Asked Questions (FAQ)

What is the general solution for -6tan^2(θ)+11tan(θ)-6=-2tan(θ) ?
The general solution for -6tan^2(θ)+11tan(θ)-6=-2tan(θ) is θ=0.58800…+pin,θ=0.98279…+pin