What is the general solution for 5tan^2(θ)-tan(θ)+12=8tan(θ)+8 ?

The general solution for 5tan^2(θ)-tan(θ)+12=8tan(θ)+8 is θ= pi/4+pin,θ=0.67474…+pin

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5tan^2(θ)-tan(θ)+12=8tan(θ)+8

Solution

5 tan^{2} (θ) - tan (θ) + 12 = 8 tan (θ) + 8

Solution

θ = \frac{π}{4} + πn, θ = 0.67474 \dots + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 38.65980 \dots^{\circ} + 18 0^{\circ} n

Solution steps

5 tan^{2} (θ) - tan (θ) + 12 = 8 tan (θ) + 8

Solve by substitution

5 tan^{2} (θ) - tan (θ) + 12 = 8 tan (θ) + 8

Let:

tan (θ) = u

5 u^{2} - u + 12 = 8 u + 8

5 u^{2} - u + 12 = 8 u + 8 : u = 1, u = \frac{4}{5}

5 u^{2} - u + 12 = 8 u + 8

Move

8

to the left side

5 u^{2} - u + 12 = 8 u + 8

Subtract

8

from both sides

5 u^{2} - u + 12 - 8 = 8 u + 8 - 8

Simplify

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

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

Move

8 u

to the left side

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

Subtract

8 u

from both sides

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

Simplify

5 u^{2} - 9 u + 4 = 0

5 u^{2} - 9 u + 4 = 0

Solve with the quadratic formula

5 u^{2} - 9 u + 4 = 0

Quadratic Equation Formula:

For

a = 5, b = - 9, c = 4

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

(- 9)^{2} - 4 \cdot 5 \cdot 4 = 1

(- 9)^{2} - 4 \cdot 5 \cdot 4

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 5 \cdot 4

Multiply the numbers:

4 \cdot 5 \cdot 4 = 80

= 9^{2} - 80

9^{2} = 81

= 81 - 80

Subtract the numbers:

81 - 80 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 9 ) \pm 1}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 1}{2 \cdot 5}, u_{2} = \frac{- ( - 9 ) - 1}{2 \cdot 5}

u = \frac{- ( - 9 ) + 1}{2 \cdot 5} : 1

\frac{- ( - 9 ) + 1}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{9 + 1}{2 \cdot 5}

Add the numbers:

9 + 1 = 10

= \frac{10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 9 ) - 1}{2 \cdot 5} : \frac{4}{5}

\frac{- ( - 9 ) - 1}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{9 - 1}{2 \cdot 5}

Subtract the numbers:

9 - 1 = 8

= \frac{8}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{8}{10}

Cancel the common factor:

2

= \frac{4}{5}

The solutions to the quadratic equation are:

u = 1, u = \frac{4}{5}

Substitute back

u = tan (θ)

tan (θ) = 1, tan (θ) = \frac{4}{5}

tan (θ) = 1, tan (θ) = \frac{4}{5}

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

tan (θ) = \frac{4}{5} : θ = arctan (\frac{4}{5}) + πn

tan (θ) = \frac{4}{5}

Apply trig inverse properties

tan (θ) = \frac{4}{5}

General solutions for

tan (θ) = \frac{4}{5}

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

θ = arctan (\frac{4}{5}) + πn

θ = arctan (\frac{4}{5}) + πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = arctan (\frac{4}{5}) + πn

Show solutions in decimal form

θ = \frac{π}{4} + πn, θ = 0.67474 \dots + πn

Popular Examples

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16 sin^{2} (θ) - 4 = 0

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cos^2(x)+2sin(x)=0

cos^{2} (x) + 2 sin (x) = 0

cos(3θ)= 1/2

cos (3 θ) = \frac{1}{2}

sin(3x+5)=cos(4x+1)

sin (3 x + 5) = cos (4 x + 1)

Frequently Asked Questions (FAQ)

What is the general solution for 5tan^2(θ)-tan(θ)+12=8tan(θ)+8 ?
The general solution for 5tan^2(θ)-tan(θ)+12=8tan(θ)+8 is θ= pi/4+pin,θ=0.67474…+pin