What is the general solution for -4tan^2(θ)-4tan(θ)-9=8tan(θ)-4 ?

The general solution for -4tan^2(θ)-4tan(θ)-9=8tan(θ)-4 is θ=-1.19028…+pin,θ=-0.46364…+pin

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Popular >

-4tan^2(θ)-4tan(θ)-9=8tan(θ)-4

Solution

- 4 tan^{2} (θ) - 4 tan (θ) - 9 = 8 tan (θ) - 4

Solution

θ = - 1.19028 \dots + πn, θ = - 0.46364 \dots + πn

Degrees

θ = - 68.19859 \dots^{\circ} + 18 0^{\circ} n, θ = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

- 4 tan^{2} (θ) - 4 tan (θ) - 9 = 8 tan (θ) - 4

Solve by substitution

- 4 tan^{2} (θ) - 4 tan (θ) - 9 = 8 tan (θ) - 4

Let:

tan (θ) = u

- 4 u^{2} - 4 u - 9 = 8 u - 4

- 4 u^{2} - 4 u - 9 = 8 u - 4 : u = - \frac{5}{2}, u = - \frac{1}{2}

- 4 u^{2} - 4 u - 9 = 8 u - 4

Move

4

to the left side

- 4 u^{2} - 4 u - 9 = 8 u - 4

Add

4

to both sides

- 4 u^{2} - 4 u - 9 + 4 = 8 u - 4 + 4

Simplify

- 4 u^{2} - 4 u - 5 = 8 u

- 4 u^{2} - 4 u - 5 = 8 u

Move

8 u

to the left side

- 4 u^{2} - 4 u - 5 = 8 u

Subtract

8 u

from both sides

- 4 u^{2} - 4 u - 5 - 8 u = 8 u - 8 u

Simplify

- 4 u^{2} - 12 u - 5 = 0

- 4 u^{2} - 12 u - 5 = 0

Solve with the quadratic formula

- 4 u^{2} - 12 u - 5 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 12, c = - 5

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 4 ) ( - 5 )}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 4 ) ( - 5 )}{2 ( - 4 )}

(- 12)^{2} - 4 (- 4) (- 5) = 8

(- 12)^{2} - 4 (- 4) (- 5)

Apply rule

- (- a) = a

= (- 12)^{2} - 4 \cdot 4 \cdot 5

Apply exponent rule:

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

n

is even

(- 12)^{2} = 1 2^{2}

= 1 2^{2} - 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 1 2^{2} - 80

1 2^{2} = 144

= 144 - 80

Subtract the numbers:

144 - 80 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 12 ) \pm 8}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 12 ) + 8}{2 ( - 4 )}, u_{2} = \frac{- ( - 12 ) - 8}{2 ( - 4 )}

u = \frac{- ( - 12 ) + 8}{2 ( - 4 )} : - \frac{5}{2}

\frac{- ( - 12 ) + 8}{2 ( - 4 )}

Remove parentheses:

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

= \frac{12 + 8}{- 2 \cdot 4}

Add the numbers:

12 + 8 = 20

= \frac{20}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{20}{- 8}

Apply the fraction rule:

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

= - \frac{20}{8}

Cancel the common factor:

4

= - \frac{5}{2}

u = \frac{- ( - 12 ) - 8}{2 ( - 4 )} : - \frac{1}{2}

\frac{- ( - 12 ) - 8}{2 ( - 4 )}

Remove parentheses:

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

= \frac{12 - 8}{- 2 \cdot 4}

Subtract the numbers:

12 - 8 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{- 8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = - \frac{5}{2}, u = - \frac{1}{2}

Substitute back

u = tan (θ)

tan (θ) = - \frac{5}{2}, tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{5}{2}, tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{5}{2} : θ = arctan (- \frac{5}{2}) + πn

tan (θ) = - \frac{5}{2}

Apply trig inverse properties

tan (θ) = - \frac{5}{2}

General solutions for

tan (θ) = - \frac{5}{2}

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

θ = arctan (- \frac{5}{2}) + πn

θ = arctan (- \frac{5}{2}) + 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{5}{2}) + πn, θ = arctan (- \frac{1}{2}) + πn

Show solutions in decimal form

θ = - 1.19028 \dots + πn, θ = - 0.46364 \dots + πn

Popular Examples

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Frequently Asked Questions (FAQ)

What is the general solution for -4tan^2(θ)-4tan(θ)-9=8tan(θ)-4 ?
The general solution for -4tan^2(θ)-4tan(θ)-9=8tan(θ)-4 is θ=-1.19028…+pin,θ=-0.46364…+pin