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

The general solution for tan^2(θ)+4tan(θ)=3 is θ=0.57338…+pin,θ=-1.35878…+pin

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Popular Trigonometry >

tan^2(θ)+4tan(θ)=3

Solution

tan^{2} (θ) + 4 tan (θ) = 3

Solution

θ = 0.57338 \dots + πn, θ = - 1.35878 \dots + πn

Degrees

θ = 32.85240 \dots^{\circ} + 18 0^{\circ} n, θ = - 77.85240 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 4 tan (θ) = 3

Solve by substitution

tan^{2} (θ) + 4 tan (θ) = 3

Let:

tan (θ) = u

u^{2} + 4 u = 3

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

u^{2} + 4 u = 3

Move

3

to the left side

u^{2} + 4 u = 3

Subtract

3

from both sides

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

Simplify

u^{2} + 4 u - 3 = 0

u^{2} + 4 u - 3 = 0

Solve with the quadratic formula

u^{2} + 4 u - 3 = 0

Quadratic Equation Formula:

For

a = 1, b = 4, c = - 3

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

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

4^{2} - 4 \cdot 1 \cdot (- 3) = 27

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

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 4^{2} + 12

4^{2} = 16

= 16 + 12

Add the numbers:

16 + 12 = 28

= 28

Prime factorization of

28 : 2^{2} \cdot 7

28

28

divides by

228 = 14 \cdot 2

= 2 \cdot 14

14

divides by

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

Apply radical rule:

= 7 2^{2}

Apply radical rule:

2^{2} = 2

= 27

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

Separate the solutions

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

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

\frac{- 4 + 2 7}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 + 2 7}{2}

Factor

- 4 + 27 : 2 (- 2 + 7)

- 4 + 27

Rewrite as

= - 2 \cdot 2 + 27

Factor out common term

2

= 2 (- 2 + 7)

= \frac{2 ( - 2 + 7 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 2 + 7

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

\frac{- 4 - 2 7}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 - 2 7}{2}

Factor

- 4 - 27 : - 2 (2 + 7)

- 4 - 27

Rewrite as

= - 2 \cdot 2 - 27

Factor out common term

2

= - 2 (2 + 7)

= - \frac{2 ( 2 + 7 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (2 + 7)

Negate

- (2 + 7) = - 2 - 7

= - 2 - 7

The solutions to the quadratic equation are:

u = - 2 + 7, u = - 2 - 7

Substitute back

u = tan (θ)

tan (θ) = - 2 + 7, tan (θ) = - 2 - 7

tan (θ) = - 2 + 7, tan (θ) = - 2 - 7

tan (θ) = - 2 + 7 : θ = arctan (- 2 + 7) + πn

tan (θ) = - 2 + 7

Apply trig inverse properties

tan (θ) = - 2 + 7

General solutions for

tan (θ) = - 2 + 7

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

θ = arctan (- 2 + 7) + πn

θ = arctan (- 2 + 7) + πn

tan (θ) = - 2 - 7 : θ = arctan (- 2 - 7) + πn

tan (θ) = - 2 - 7

Apply trig inverse properties

tan (θ) = - 2 - 7

General solutions for

tan (θ) = - 2 - 7

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

θ = arctan (- 2 - 7) + πn

θ = arctan (- 2 - 7) + πn

Combine all the solutions

θ = arctan (- 2 + 7) + πn, θ = arctan (- 2 - 7) + πn

Show solutions in decimal form

θ = 0.57338 \dots + πn, θ = - 1.35878 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+4tan(θ)=3 ?
The general solution for tan^2(θ)+4tan(θ)=3 is θ=0.57338…+pin,θ=-1.35878…+pin