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

The general solution for 2tan^2(θ)+3tan(θ)+4=3 is θ=-0.46364…+pin,θ=(3pi)/4+pin

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2tan^2(θ)+3tan(θ)+4=3

Solution

2 tan^{2} (θ) + 3 tan (θ) + 4 = 3

Solution

θ = - 0.46364 \dots + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = - 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

2 tan^{2} (θ) + 3 tan (θ) + 4 = 3

Solve by substitution

2 tan^{2} (θ) + 3 tan (θ) + 4 = 3

Let:

tan (θ) = u

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

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

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

Move

3

to the left side

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

Subtract

3

from both sides

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

Simplify

2 u^{2} + 3 u + 1 = 0

2 u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = 1

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

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

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

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- 3 + 1}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

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

\frac{- 3 - 1}{2 \cdot 2}

Subtract the numbers:

- 3 - 1 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = tan (θ)

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

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

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

tan (θ) = - 1 : θ = \frac{3 π}{4} + πn

tan (θ) = - 1

General solutions for

tan (θ) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = arctan (- \frac{1}{2}) + πn, θ = \frac{3 π}{4} + πn

Show solutions in decimal form

θ = - 0.46364 \dots + πn, θ = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(θ)+3tan(θ)+4=3 ?
The general solution for 2tan^2(θ)+3tan(θ)+4=3 is θ=-0.46364…+pin,θ=(3pi)/4+pin