What is the general solution for 3tan^2(θ)+7tan(θ)-2=-4 ?

The general solution for 3tan^2(θ)+7tan(θ)-2=-4 is θ=-0.32175…+pin,θ=-1.10714…+pin

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3tan^2(θ)+7tan(θ)-2=-4

Solution

3 tan^{2} (θ) + 7 tan (θ) - 2 = - 4

Solution

θ = - 0.32175 \dots + πn, θ = - 1.10714 \dots + πn

Degrees

θ = - 18.43494 \dots^{\circ} + 18 0^{\circ} n, θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 tan^{2} (θ) + 7 tan (θ) - 2 = - 4

Solve by substitution

3 tan^{2} (θ) + 7 tan (θ) - 2 = - 4

Let:

tan (θ) = u

3 u^{2} + 7 u - 2 = - 4

3 u^{2} + 7 u - 2 = - 4 : u = - \frac{1}{3}, u = - 2

3 u^{2} + 7 u - 2 = - 4

Move

4

to the left side

3 u^{2} + 7 u - 2 = - 4

Add

4

to both sides

3 u^{2} + 7 u - 2 + 4 = - 4 + 4

Simplify

3 u^{2} + 7 u + 2 = 0

3 u^{2} + 7 u + 2 = 0

Solve with the quadratic formula

3 u^{2} + 7 u + 2 = 0

Quadratic Equation Formula:

For

a = 3, b = 7, c = 2

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

7^{2} - 4 \cdot 3 \cdot 2 = 5

7^{2} - 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 7 \pm 5}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 7 + 5}{2 \cdot 3}, u_{2} = \frac{- 7 - 5}{2 \cdot 3}

u = \frac{- 7 + 5}{2 \cdot 3} : - \frac{1}{3}

\frac{- 7 + 5}{2 \cdot 3}

Add/Subtract the numbers:

- 7 + 5 = - 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{6}

Apply the fraction rule:

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

= - \frac{2}{6}

Cancel the common factor:

2

= - \frac{1}{3}

u = \frac{- 7 - 5}{2 \cdot 3} : - 2

\frac{- 7 - 5}{2 \cdot 3}

Subtract the numbers:

- 7 - 5 = - 12

= \frac{- 12}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 12}{6}

Apply the fraction rule:

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

= - \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = tan (θ)

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

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

Show solutions in decimal form

θ = - 0.32175 \dots + πn, θ = - 1.10714 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(θ)+7tan(θ)-2=-4 ?
The general solution for 3tan^2(θ)+7tan(θ)-2=-4 is θ=-0.32175…+pin,θ=-1.10714…+pin