What is the general solution for tan^2(x)+tan(x)=0,0<= x<2pi ?

Q: What is the general solution for tan^2(x)+tan(x)=0,0<= x<2pi ?

The general solution for tan^2(x)+tan(x)=0,0<= x<2pi is x=0,x=pi,x=(3pi)/4 ,x=(7pi)/4

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tan^2(x)+tan(x)=0,0<= x<2pi

Solution

tan^{2} (x) + tan (x) = 0, 0 \leq x < 2 π

Solution

x = 0, x = π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

Solution steps

tan^{2} (x) + tan (x) = 0, 0 \leq x < 2 π

Solve by substitution

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

Let:

tan (x) = 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 (x)

tan (x) = 0, tan (x) = - 1

tan (x) = 0, tan (x) = - 1

tan (x) = 0, 0 \leq x < 2 π : x = 0, x = π

tan (x) = 0, 0 \leq x < 2 π

General solutions for

tan (x) = 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}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

tan (x) = - 1, 0 \leq x < 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) = - 1, 0 \leq x < 2 π

General solutions for

tan (x) = - 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}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = 0, x = π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

Popular Examples

csc(x)-1=0

csc (x) - 1 = 0

cos(2x)+5cos(x)+3=0

cos (2 x) + 5 cos (x) + 3 = 0

2cos(θ)=1

2 cos (θ) = 1

sin(x+pi/6)-sin(x-pi/6)= 1/2

sin (x + \frac{π}{6}) - sin (x - \frac{π}{6}) = \frac{1}{2}

tan(θ)= 5/12

tan (θ) = \frac{5}{12}

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+tan(x)=0,0<= x<2pi ?
The general solution for tan^2(x)+tan(x)=0,0<= x<2pi is x=0,x=pi,x=(3pi)/4 ,x=(7pi)/4