What is the general solution for 2tan^2(θ)+tan(θ)=0 ?

The general solution for 2tan^2(θ)+tan(θ)=0 is θ=pin,θ=-0.46364…+pin

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Popular >

2tan^2(θ)+tan(θ)=0

Solution

2 tan^{2} (θ) + tan (θ) = 0

Solution

θ = πn, θ = - 0.46364 \dots + πn

Degrees

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

Solution steps

2 tan^{2} (θ) + tan (θ) = 0

Solve by substitution

2 tan^{2} (θ) + tan (θ) = 0

Let:

tan (θ) = u

2 u^{2} + u = 0

2 u^{2} + u = 0 : u = 0, u = - \frac{1}{2}

2 u^{2} + u = 0

Solve with the quadratic formula

2 u^{2} + u = 0

Quadratic Equation Formula:

For

a = 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1^{2} - 4 \cdot 2 \cdot 0 = 1

1^{2} - 4 \cdot 2 \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- 1 + 1}{2 \cdot 2} : 0

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

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

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

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

Subtract the numbers:

- 1 - 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}

The solutions to the quadratic equation are:

u = 0, u = - \frac{1}{2}

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - \frac{1}{2}

tan (θ) = 0, tan (θ) = - \frac{1}{2}

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 0

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 = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

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

Combine all the solutions

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

Show solutions in decimal form

θ = πn, θ = - 0.46364 \dots + πn

Popular Examples

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2 cos^{2} (θ) + sin (θ) = 1

tan(2x)-sqrt(3)=0

tan (2 x) - 3 = 0

0=cos(2x)

0 = cos (2 x)

cos(5x)=0

cos (5 x) = 0

tan(2x)+tan(x)=0

tan (2 x) + tan (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(θ)+tan(θ)=0 ?
The general solution for 2tan^2(θ)+tan(θ)=0 is θ=pin,θ=-0.46364…+pin