What is the general solution for 4tan^2(θ)+16tan(θ)=7tan(θ)-2 ?

The general solution for 4tan^2(θ)+16tan(θ)=7tan(θ)-2 is θ=-0.24497…+pin,θ=-1.10714…+pin

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Popular >

4tan^2(θ)+16tan(θ)=7tan(θ)-2

Solution

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

Solution

θ = - 0.24497 \dots + πn, θ = - 1.10714 \dots + πn

Degrees

θ = - 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

Solve by substitution

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

Let:

tan (θ) = u

4 u^{2} + 16 u = 7 u - 2

4 u^{2} + 16 u = 7 u - 2 : u = - \frac{1}{4}, u = - 2

4 u^{2} + 16 u = 7 u - 2

Move

2

to the left side

4 u^{2} + 16 u = 7 u - 2

Add

2

to both sides

4 u^{2} + 16 u + 2 = 7 u - 2 + 2

Simplify

4 u^{2} + 16 u + 2 = 7 u

4 u^{2} + 16 u + 2 = 7 u

Move

7 u

to the left side

4 u^{2} + 16 u + 2 = 7 u

Subtract

7 u

from both sides

4 u^{2} + 16 u + 2 - 7 u = 7 u - 7 u

Simplify

4 u^{2} + 9 u + 2 = 0

4 u^{2} + 9 u + 2 = 0

Solve with the quadratic formula

4 u^{2} + 9 u + 2 = 0

Quadratic Equation Formula:

For

a = 4, b = 9, c = 2

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

9^{2} - 4 \cdot 4 \cdot 2 = 7

9^{2} - 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

Subtract the numbers:

81 - 32 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 9 \pm 7}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 9 + 7}{2 \cdot 4}, u_{2} = \frac{- 9 - 7}{2 \cdot 4}

u = \frac{- 9 + 7}{2 \cdot 4} : - \frac{1}{4}

\frac{- 9 + 7}{2 \cdot 4}

Add/Subtract the numbers:

- 9 + 7 = - 2

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2}{8}

Apply the fraction rule:

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

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

u = \frac{- 9 - 7}{2 \cdot 4} : - 2

\frac{- 9 - 7}{2 \cdot 4}

Subtract the numbers:

- 9 - 7 = - 16

= \frac{- 16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 16}{8}

Apply the fraction rule:

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

= - \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= - 2

The solutions to the quadratic equation are:

u = - \frac{1}{4}, u = - 2

Substitute back

u = tan (θ)

tan (θ) = - \frac{1}{4}, tan (θ) = - 2

tan (θ) = - \frac{1}{4}, tan (θ) = - 2

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

tan (θ) = - \frac{1}{4}

Apply trig inverse properties

tan (θ) = - \frac{1}{4}

General solutions for

tan (θ) = - \frac{1}{4}

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

θ = arctan (- \frac{1}{4}) + πn

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

Show solutions in decimal form

θ = - 0.24497 \dots + πn, θ = - 1.10714 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4tan^2(θ)+16tan(θ)=7tan(θ)-2 ?
The general solution for 4tan^2(θ)+16tan(θ)=7tan(θ)-2 is θ=-0.24497…+pin,θ=-1.10714…+pin