What is the general solution for -9tan^2(θ)-1=-6tan(θ) ?

The general solution for -9tan^2(θ)-1=-6tan(θ) is θ=0.32175…+pin

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Popular >

-9tan^2(θ)-1=-6tan(θ)

Solution

- 9 tan^{2} (θ) - 1 = - 6 tan (θ)

Solution

θ = 0.32175 \dots + πn

Degrees

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

Solution steps

- 9 tan^{2} (θ) - 1 = - 6 tan (θ)

Solve by substitution

- 9 tan^{2} (θ) - 1 = - 6 tan (θ)

Let:

tan (θ) = u

- 9 u^{2} - 1 = - 6 u

- 9 u^{2} - 1 = - 6 u : u = \frac{1}{3}

- 9 u^{2} - 1 = - 6 u

Move

6 u

to the left side

- 9 u^{2} - 1 = - 6 u

Add

6 u

to both sides

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

Simplify

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 9, b = 6, c = - 1

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 9 ) ( - 1 )}{2 ( - 9 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 9 ) ( - 1 )}{2 ( - 9 )}

6^{2} - 4 (- 9) (- 1) = 0

6^{2} - 4 (- 9) (- 1)

Apply rule

- (- a) = a

= 6^{2} - 4 \cdot 9 \cdot 1

Multiply the numbers:

4 \cdot 9 \cdot 1 = 36

= 6^{2} - 36

6^{2} = 36

= 36 - 36

Subtract the numbers:

36 - 36 = 0

= 0

u_{1, 2} = \frac{- 6 \pm 0}{2 ( - 9 )}

u = \frac{- 6}{2 ( - 9 )}

\frac{- 6}{2 ( - 9 )} = \frac{1}{3}

\frac{- 6}{2 ( - 9 )}

Remove parentheses:

(- a) = - a

= \frac{- 6}{- 2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 6}{- 18}

Apply the fraction rule:

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

= \frac{6}{18}

Cancel the common factor:

6

= \frac{1}{3}

u = \frac{1}{3}

The solution to the quadratic equation is:

u = \frac{1}{3}

Substitute back

u = tan (θ)

tan (θ) = \frac{1}{3}

tan (θ) = \frac{1}{3}

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

Combine all the solutions

θ = arctan (\frac{1}{3}) + πn

Show solutions in decimal form

θ = 0.32175 \dots + πn

Popular Examples

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3tan^2(θ)=1

3 tan^{2} (θ) = 1

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3 cos (x) = 2 sin^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for -9tan^2(θ)-1=-6tan(θ) ?
The general solution for -9tan^2(θ)-1=-6tan(θ) is θ=0.32175…+pin