What is the general solution for -8tan^2(θ)-4tan(θ)-11=6tan(θ)-8 ?

The general solution for -8tan^2(θ)-4tan(θ)-11=6tan(θ)-8 is θ=-0.64350…+pin,θ=-0.46364…+pin

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Popular >

-8tan^2(θ)-4tan(θ)-11=6tan(θ)-8

Solution

- 8 tan^{2} (θ) - 4 tan (θ) - 11 = 6 tan (θ) - 8

Solution

θ = - 0.64350 \dots + πn, θ = - 0.46364 \dots + πn

Degrees

θ = - 36.86989 \dots^{\circ} + 18 0^{\circ} n, θ = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

- 8 tan^{2} (θ) - 4 tan (θ) - 11 = 6 tan (θ) - 8

Solve by substitution

- 8 tan^{2} (θ) - 4 tan (θ) - 11 = 6 tan (θ) - 8

Let:

tan (θ) = u

- 8 u^{2} - 4 u - 11 = 6 u - 8

- 8 u^{2} - 4 u - 11 = 6 u - 8 : u = - \frac{3}{4}, u = - \frac{1}{2}

- 8 u^{2} - 4 u - 11 = 6 u - 8

Move

8

to the left side

- 8 u^{2} - 4 u - 11 = 6 u - 8

Add

8

to both sides

- 8 u^{2} - 4 u - 11 + 8 = 6 u - 8 + 8

Simplify

- 8 u^{2} - 4 u - 3 = 6 u

- 8 u^{2} - 4 u - 3 = 6 u

Move

6 u

to the left side

- 8 u^{2} - 4 u - 3 = 6 u

Subtract

6 u

from both sides

- 8 u^{2} - 4 u - 3 - 6 u = 6 u - 6 u

Simplify

- 8 u^{2} - 10 u - 3 = 0

- 8 u^{2} - 10 u - 3 = 0

Solve with the quadratic formula

- 8 u^{2} - 10 u - 3 = 0

Quadratic Equation Formula:

For

a = - 8, b = - 10, c = - 3

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 8 ) ( - 3 )}{2 ( - 8 )}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 8 ) ( - 3 )}{2 ( - 8 )}

(- 10)^{2} - 4 (- 8) (- 3) = 2

(- 10)^{2} - 4 (- 8) (- 3)

Apply rule

- (- a) = a

= (- 10)^{2} - 4 \cdot 8 \cdot 3

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 8 \cdot 3

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 1 0^{2} - 96

1 0^{2} = 100

= 100 - 96

Subtract the numbers:

100 - 96 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- ( - 10 ) + 2}{2 ( - 8 )}, u_{2} = \frac{- ( - 10 ) - 2}{2 ( - 8 )}

u = \frac{- ( - 10 ) + 2}{2 ( - 8 )} : - \frac{3}{4}

\frac{- ( - 10 ) + 2}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{10 + 2}{- 2 \cdot 8}

Add the numbers:

10 + 2 = 12

= \frac{12}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{12}{- 16}

Apply the fraction rule:

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

= - \frac{12}{16}

Cancel the common factor:

4

= - \frac{3}{4}

u = \frac{- ( - 10 ) - 2}{2 ( - 8 )} : - \frac{1}{2}

\frac{- ( - 10 ) - 2}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{10 - 2}{- 2 \cdot 8}

Subtract the numbers:

10 - 2 = 8

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8}{- 16}

Apply the fraction rule:

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

= - \frac{8}{16}

Cancel the common factor:

8

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = - \frac{3}{4}, u = - \frac{1}{2}

Substitute back

u = tan (θ)

tan (θ) = - \frac{3}{4}, tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{3}{4}, tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{3}{4} : θ = arctan (- \frac{3}{4}) + πn

tan (θ) = - \frac{3}{4}

Apply trig inverse properties

tan (θ) = - \frac{3}{4}

General solutions for

tan (θ) = - \frac{3}{4}

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

θ = arctan (- \frac{3}{4}) + πn

θ = arctan (- \frac{3}{4}) + πn

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

Combine all the solutions

θ = arctan (- \frac{3}{4}) + πn, θ = arctan (- \frac{1}{2}) + πn

Show solutions in decimal form

θ = - 0.64350 \dots + πn, θ = - 0.46364 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for -8tan^2(θ)-4tan(θ)-11=6tan(θ)-8 ?
The general solution for -8tan^2(θ)-4tan(θ)-11=6tan(θ)-8 is θ=-0.64350…+pin,θ=-0.46364…+pin