What is the general solution for tan^2(x)+tan(x)-12=0 ?

The general solution for tan^2(x)+tan(x)-12=0 is x=1.24904…+pin,x=-1.32581…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(x)+tan(x)-12=0

Solution

tan^{2} (x) + tan (x) - 12 = 0

Solution

x = 1.24904 \dots + πn, x = - 1.32581 \dots + πn

Degrees

x = 71.56505 \dots^{\circ} + 18 0^{\circ} n, x = - 75.96375 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) + tan (x) - 12 = 0

Solve by substitution

tan^{2} (x) + tan (x) - 12 = 0

Let:

tan (x) = u

u^{2} + u - 12 = 0

u^{2} + u - 12 = 0 : u = 3, u = - 4

u^{2} + u - 12 = 0

Solve with the quadratic formula

u^{2} + u - 12 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 12

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

1^{2} - 4 \cdot 1 \cdot (- 12)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 12)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 12

Multiply the numbers:

4 \cdot 1 \cdot 12 = 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 7}{2 \cdot 1}, u_{2} = \frac{- 1 - 7}{2 \cdot 1}

u = \frac{- 1 + 7}{2 \cdot 1} : 3

\frac{- 1 + 7}{2 \cdot 1}

Add/Subtract the numbers:

- 1 + 7 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

u = \frac{- 1 - 7}{2 \cdot 1} : - 4

\frac{- 1 - 7}{2 \cdot 1}

Subtract the numbers:

- 1 - 7 = - 8

= \frac{- 8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 8}{2}

Apply the fraction rule:

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

= - \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= - 4

The solutions to the quadratic equation are:

u = 3, u = - 4

Substitute back

u = tan (x)

tan (x) = 3, tan (x) = - 4

tan (x) = 3, tan (x) = - 4

tan (x) = 3 : x = arctan (3) + πn

tan (x) = 3

Apply trig inverse properties

tan (x) = 3

General solutions for

tan (x) = 3

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

x = arctan (3) + πn

x = arctan (3) + πn

tan (x) = - 4 : x = arctan (- 4) + πn

tan (x) = - 4

Apply trig inverse properties

tan (x) = - 4

General solutions for

tan (x) = - 4

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

x = arctan (- 4) + πn

x = arctan (- 4) + πn

Combine all the solutions

x = arctan (3) + πn, x = arctan (- 4) + πn

Show solutions in decimal form

x = 1.24904 \dots + πn, x = - 1.32581 \dots + πn

Popular Examples

sin(x)+1=cos(x)

sin (x) + 1 = cos (x)

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

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

sec^2(x)=0

sec^{2} (x) = 0

sec^2(x)+tan(x)=1

sec^{2} (x) + tan (x) = 1

4sin^2(θ)-1=0

4 sin^{2} (θ) - 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+tan(x)-12=0 ?
The general solution for tan^2(x)+tan(x)-12=0 is x=1.24904…+pin,x=-1.32581…+pin