What is the general solution for tan^2(θ)+5tan(θ)=0 ?

The general solution for tan^2(θ)+5tan(θ)=0 is θ=pin,θ=-1.37340…+pin

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Popular >

tan^2(θ)+5tan(θ)=0

Solution

tan^{2} (θ) + 5 tan (θ) = 0

Solution

θ = πn, θ = - 1.37340 \dots + πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = - 78.69006 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) + 5 tan (θ) = 0

Solve by substitution

tan^{2} (θ) + 5 tan (θ) = 0

Let:

tan (θ) = u

u^{2} + 5 u = 0

u^{2} + 5 u = 0 : u = 0, u = - 5

u^{2} + 5 u = 0

Solve with the quadratic formula

u^{2} + 5 u = 0

Quadratic Equation Formula:

For

a = 1, b = 5, c = 0

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 0 = 5

5^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

Apply radical rule:

assuming

a \geq 0

= 5

u_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 5}{2 \cdot 1}, u_{2} = \frac{- 5 - 5}{2 \cdot 1}

u = \frac{- 5 + 5}{2 \cdot 1} : 0

\frac{- 5 + 5}{2 \cdot 1}

Add/Subtract the numbers:

- 5 + 5 = 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{- 5 - 5}{2 \cdot 1} : - 5

\frac{- 5 - 5}{2 \cdot 1}

Subtract the numbers:

- 5 - 5 = - 10

= \frac{- 10}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 10}{2}

Apply the fraction rule:

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

= - \frac{10}{2}

Divide the numbers:

\frac{10}{2} = 5

= - 5

The solutions to the quadratic equation are:

u = 0, u = - 5

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - 5

tan (θ) = 0, tan (θ) = - 5

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) = - 5 : θ = arctan (- 5) + πn

tan (θ) = - 5

Apply trig inverse properties

tan (θ) = - 5

General solutions for

tan (θ) = - 5

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

θ = arctan (- 5) + πn

θ = arctan (- 5) + πn

Combine all the solutions

θ = πn, θ = arctan (- 5) + πn

Show solutions in decimal form

θ = πn, θ = - 1.37340 \dots + πn

Popular Examples

sin(4x)-sin(6x)=0

sin (4 x) - sin (6 x) = 0

2=sec(x)+sec^2(x)

2 = sec (x) + sec^{2} (x)

4cos(θ)-5=0

4 cos (θ) - 5 = 0

cos(u)=-24/25

cos (u) = - \frac{24}{25}

4sin(y)= 1/(cos(y))

4 sin (y) = \frac{1}{cos ( y )}

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)+5tan(θ)=0 ?
The general solution for tan^2(θ)+5tan(θ)=0 is θ=pin,θ=-1.37340…+pin