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

The general solution for tan^2(θ)+2tan(θ)=0,tan(2θ)+2tan(θ)=0 is No Solution for θ\in\mathbb{R}

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tan^2(θ)+2tan(θ)=0,tan(2θ)+2tan(θ)=0

Solution

tan^{2} (θ) + 2 tan (θ) = 0, tan (2 θ) + 2 tan (θ) = 0

Solution

No Solution for θ \in R

Solution steps

tan^{2} (θ) + 2 tan (θ) = 0, tan (2 θ) + 2 tan (θ) = 0

Solve by substitution

tan^{2} (θ) + 2 tan (θ) = 0

Let:

tan (θ) = u

u^{2} + 2 u = 0

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

u^{2} + 2 u = 0

Solve with the quadratic formula

u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 0

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

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

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

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

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 2 + 2 = 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{- 2 - 2}{2 \cdot 1} : - 2

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

Subtract the numbers:

- 2 - 2 = - 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = 0, u = - 2

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - 2

tan (θ) = 0, tan (θ) = - 2

tan (θ) = 0, tan (2 θ) + 2 tan (θ) = 0 :

No Solution

tan (θ) = 0, tan (2 θ) + 2 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

Solutions for the range

tan (2 θ) + 2 tan (θ) = 0

No Solution

tan (θ) = - 2, tan (2 θ) + 2 tan (θ) = 0 :

No Solution

tan (θ) = - 2, tan (2 θ) + 2 tan (θ) = 0

Apply trig inverse properties

tan (θ) = - 2

General solutions for

tan (θ) = - 2

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

θ = arctan (- 2) + πn

θ = arctan (- 2) + πn

Solutions for the range

tan (2 θ) + 2 tan (θ) = 0

No Solution

Combine all the solutions

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+2tan(θ)=0,tan(2θ)+2tan(θ)=0 ?
The general solution for tan^2(θ)+2tan(θ)=0,tan(2θ)+2tan(θ)=0 is No Solution for θ\in\mathbb{R}