What is the general solution for tan^2(x)+4tan(x)-5=0 ?

The general solution for tan^2(x)+4tan(x)-5=0 is x= pi/4+pin,x=-1.37340…+pin

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tan^2(x)+4tan(x)-5=0

Solution

tan^{2} (x) + 4 tan (x) - 5 = 0

Solution

x = \frac{π}{4} + πn, x = - 1.37340 \dots + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 78.69006 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) + 4 tan (x) - 5 = 0

Solve by substitution

tan^{2} (x) + 4 tan (x) - 5 = 0

Let:

tan (x) = u

u^{2} + 4 u - 5 = 0

u^{2} + 4 u - 5 = 0 : u = 1, u = - 5

u^{2} + 4 u - 5 = 0

Solve with the quadratic formula

u^{2} + 4 u - 5 = 0

Quadratic Equation Formula:

For

a = 1, b = 4, c = - 5

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

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

4^{2} - 4 \cdot 1 \cdot (- 5) = 6

4^{2} - 4 \cdot 1 \cdot (- 5)

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 1 \cdot 5

Multiply the numbers:

4 \cdot 1 \cdot 5 = 20

= 4^{2} + 20

4^{2} = 16

= 16 + 20

Add the numbers:

16 + 20 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- 4 \pm 6}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 4 + 6}{2 \cdot 1}, u_{2} = \frac{- 4 - 6}{2 \cdot 1}

u = \frac{- 4 + 6}{2 \cdot 1} : 1

\frac{- 4 + 6}{2 \cdot 1}

Add/Subtract the numbers:

- 4 + 6 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 4 - 6}{2 \cdot 1} : - 5

\frac{- 4 - 6}{2 \cdot 1}

Subtract the numbers:

- 4 - 6 = - 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 = 1, u = - 5

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = - 5

tan (x) = 1, tan (x) = - 5

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

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{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 5 : x = arctan (- 5) + πn

tan (x) = - 5

Apply trig inverse properties

tan (x) = - 5

General solutions for

tan (x) = - 5

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

x = arctan (- 5) + πn

x = arctan (- 5) + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = arctan (- 5) + πn

Show solutions in decimal form

x = \frac{π}{4} + πn, x = - 1.37340 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+4tan(x)-5=0 ?
The general solution for tan^2(x)+4tan(x)-5=0 is x= pi/4+pin,x=-1.37340…+pin