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

The general solution for tan^2(θ)+tan(θ)=0 is θ=pin,θ=(3pi)/4+pin

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tan^2(θ)+tan(θ)=0

Solution

tan^{2} (θ) + tan (θ) = 0

Solution

θ = πn, θ = \frac{3 π}{4} + πn

Degrees

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

Solution steps

tan^{2} (θ) + tan (θ) = 0

Solve by substitution

tan^{2} (θ) + tan (θ) = 0

Let:

tan (θ) = u

u^{2} + u = 0

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

u^{2} + u = 0

Solve with the quadratic formula

u^{2} + u = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = 0

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \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 1}

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

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

= 0

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

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

Subtract the numbers:

- 1 - 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 0, u = - 1

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - 1

tan (θ) = 0, tan (θ) = - 1

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) = - 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 = πn, θ = \frac{3 π}{4} + πn

Popular Examples

sin^2(x)+sin(x)-2=0

sin^{2} (x) + sin (x) - 2 = 0

cos(2x)=cos(x)-1

cos (2 x) = cos (x) - 1

2sin^2(x)-3sin(x)-2=0

2 sin^{2} (x) - 3 sin (x) - 2 = 0

tan^2(x)+tan(x)-6=0

tan^{2} (x) + tan (x) - 6 = 0

2cos(2x)+sqrt(3)=0

2 cos (2 x) + 3 = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+tan(θ)=0 ?
The general solution for tan^2(θ)+tan(θ)=0 is θ=pin,θ=(3pi)/4+pin