What is the general solution for -2tan^2(θ)+2tan(θ)+1=3tan(θ) ?

The general solution for -2tan^2(θ)+2tan(θ)+1=3tan(θ) is θ=(3pi)/4+pin,θ=0.46364…+pin

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-2tan^2(θ)+2tan(θ)+1=3tan(θ)

Solution

- 2 tan^{2} (θ) + 2 tan (θ) + 1 = 3 tan (θ)

Solution

θ = \frac{3 π}{4} + πn, θ = 0.46364 \dots + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n, θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

- 2 tan^{2} (θ) + 2 tan (θ) + 1 = 3 tan (θ)

Solve by substitution

- 2 tan^{2} (θ) + 2 tan (θ) + 1 = 3 tan (θ)

Let:

tan (θ) = u

- 2 u^{2} + 2 u + 1 = 3 u

- 2 u^{2} + 2 u + 1 = 3 u : u = - 1, u = \frac{1}{2}

- 2 u^{2} + 2 u + 1 = 3 u

Move

3 u

to the left side

- 2 u^{2} + 2 u + 1 = 3 u

Subtract

3 u

from both sides

- 2 u^{2} + 2 u + 1 - 3 u = 3 u - 3 u

Simplify

- 2 u^{2} - u + 1 = 0

- 2 u^{2} - u + 1 = 0

Solve with the quadratic formula

- 2 u^{2} - u + 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 2) \cdot 1 = 3

(- 1)^{2} - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 3}{2 ( - 2 )} : - 1

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 3}{- 2 \cdot 2}

Add the numbers:

1 + 3 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 - 3}{- 2 \cdot 2}

Subtract the numbers:

1 - 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = tan (θ)

tan (θ) = - 1, tan (θ) = \frac{1}{2}

tan (θ) = - 1, tan (θ) = \frac{1}{2}

tan (θ) = - 1 : θ = \frac{3 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{3 π}{4} + πn

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

Show solutions in decimal form

θ = \frac{3 π}{4} + πn, θ = 0.46364 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for -2tan^2(θ)+2tan(θ)+1=3tan(θ) ?
The general solution for -2tan^2(θ)+2tan(θ)+1=3tan(θ) is θ=(3pi)/4+pin,θ=0.46364…+pin