What is the general solution for 2tan^2(θ)-5tan(θ)+4=-8tan(θ)+6 ?

The general solution for 2tan^2(θ)-5tan(θ)+4=-8tan(θ)+6 is θ=0.46364…+pin,θ=-1.10714…+pin

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Popular >

2tan^2(θ)-5tan(θ)+4=-8tan(θ)+6

Solution

2 tan^{2} (θ) - 5 tan (θ) + 4 = - 8 tan (θ) + 6

Solution

θ = 0.46364 \dots + πn, θ = - 1.10714 \dots + πn

Degrees

θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 tan^{2} (θ) - 5 tan (θ) + 4 = - 8 tan (θ) + 6

Solve by substitution

2 tan^{2} (θ) - 5 tan (θ) + 4 = - 8 tan (θ) + 6

Let:

tan (θ) = u

2 u^{2} - 5 u + 4 = - 8 u + 6

2 u^{2} - 5 u + 4 = - 8 u + 6 : u = \frac{1}{2}, u = - 2

2 u^{2} - 5 u + 4 = - 8 u + 6

Move

6

to the left side

2 u^{2} - 5 u + 4 = - 8 u + 6

Subtract

6

from both sides

2 u^{2} - 5 u + 4 - 6 = - 8 u + 6 - 6

Simplify

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

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

Move

8 u

to the left side

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

Add

8 u

to both sides

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

Simplify

2 u^{2} + 3 u - 2 = 0

2 u^{2} + 3 u - 2 = 0

Solve with the quadratic formula

2 u^{2} + 3 u - 2 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = - 2

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

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

3^{2} - 4 \cdot 2 (- 2) = 5

3^{2} - 4 \cdot 2 (- 2)

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 5}{2 \cdot 2}, u_{2} = \frac{- 3 - 5}{2 \cdot 2}

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 5 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 3 - 5}{2 \cdot 2} : - 2

\frac{- 3 - 5}{2 \cdot 2}

Subtract the numbers:

- 3 - 5 = - 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = tan (θ)

tan (θ) = \frac{1}{2}, tan (θ) = - 2

tan (θ) = \frac{1}{2}, tan (θ) = - 2

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) = - 2 : θ = arctan (- 2) + πn

tan (θ) = - 2

Apply trig inverse properties

tan (θ) = - 2

General solutions for

tan (θ) = - 2

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

θ = arctan (- 2) + πn

θ = arctan (- 2) + πn

Combine all the solutions

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

Show solutions in decimal form

θ = 0.46364 \dots + πn, θ = - 1.10714 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(θ)-5tan(θ)+4=-8tan(θ)+6 ?
The general solution for 2tan^2(θ)-5tan(θ)+4=-8tan(θ)+6 is θ=0.46364…+pin,θ=-1.10714…+pin