What is the general solution for -6tan^2(θ)+4tan(θ)=-tan(θ)+1 ?

The general solution for -6tan^2(θ)+4tan(θ)=-tan(θ)+1 is θ=0.32175…+pin,θ=0.46364…+pin

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Popular >

-6tan^2(θ)+4tan(θ)=-tan(θ)+1

Solution

- 6 tan^{2} (θ) + 4 tan (θ) = - tan (θ) + 1

Solution

θ = 0.32175 \dots + πn, θ = 0.46364 \dots + πn

Degrees

θ = 18.43494 \dots^{\circ} + 18 0^{\circ} n, θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

- 6 tan^{2} (θ) + 4 tan (θ) = - tan (θ) + 1

Solve by substitution

- 6 tan^{2} (θ) + 4 tan (θ) = - tan (θ) + 1

Let:

tan (θ) = u

- 6 u^{2} + 4 u = - u + 1

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

- 6 u^{2} + 4 u = - u + 1

Move

1

to the left side

- 6 u^{2} + 4 u = - u + 1

Subtract

1

from both sides

- 6 u^{2} + 4 u - 1 = - u + 1 - 1

Simplify

- 6 u^{2} + 4 u - 1 = - u

- 6 u^{2} + 4 u - 1 = - u

Move

u

to the left side

- 6 u^{2} + 4 u - 1 = - u

Add

u

to both sides

- 6 u^{2} + 4 u - 1 + u = - u + u

Simplify

- 6 u^{2} + 5 u - 1 = 0

- 6 u^{2} + 5 u - 1 = 0

Solve with the quadratic formula

- 6 u^{2} + 5 u - 1 = 0

Quadratic Equation Formula:

For

a = - 6, b = 5, c = - 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 6 ) ( - 1 )}{2 ( - 6 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 6 ) ( - 1 )}{2 ( - 6 )}

5^{2} - 4 (- 6) (- 1) = 1

5^{2} - 4 (- 6) (- 1)

Apply rule

- (- a) = a

= 5^{2} - 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- 5 + 1}{2 ( - 6 )}, u_{2} = \frac{- 5 - 1}{2 ( - 6 )}

u = \frac{- 5 + 1}{2 ( - 6 )} : \frac{1}{3}

\frac{- 5 + 1}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 1}{- 2 \cdot 6}

Add/Subtract the numbers:

- 5 + 1 = - 4

= \frac{- 4}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 4}{- 12}

Apply the fraction rule:

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

= \frac{4}{12}

Cancel the common factor:

4

= \frac{1}{3}

u = \frac{- 5 - 1}{2 ( - 6 )} : \frac{1}{2}

\frac{- 5 - 1}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 1}{- 2 \cdot 6}

Subtract the numbers:

- 5 - 1 = - 6

= \frac{- 6}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

Apply the fraction rule:

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

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = tan (θ)

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

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

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = \frac{1}{3}

Apply trig inverse properties

tan (θ) = \frac{1}{3}

General solutions for

tan (θ) = \frac{1}{3}

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

θ = arctan (\frac{1}{3}) + πn

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

Show solutions in decimal form

θ = 0.32175 \dots + πn, θ = 0.46364 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for -6tan^2(θ)+4tan(θ)=-tan(θ)+1 ?
The general solution for -6tan^2(θ)+4tan(θ)=-tan(θ)+1 is θ=0.32175…+pin,θ=0.46364…+pin