What is the general solution for tan^2(x)+4tan(x)+1=0 ?

The general solution for tan^2(x)+4tan(x)+1=0 is x=-0.26179…+pin,x=-1.30899…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

tan^2(x)+4tan(x)+1=0

Solution

tan^{2} (x) + 4 tan (x) + 1 = 0

Solution

x = - 0.26179 \dots + πn, x = - 1.30899 \dots + πn

Degrees

x = - 1 5^{\circ} + 18 0^{\circ} n, x = - 7 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) + 4 tan (x) + 1 = 0

Solve by substitution

tan^{2} (x) + 4 tan (x) + 1 = 0

Let:

tan (x) = u

u^{2} + 4 u + 1 = 0

u^{2} + 4 u + 1 = 0 : u = - 2 + 3, u = - 2 - 3

u^{2} + 4 u + 1 = 0

Solve with the quadratic formula

u^{2} + 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 4, c = 1

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

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

4^{2} - 4 \cdot 1 \cdot 1 = 23

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4^{2} - 4

4^{2} = 16

= 16 - 4

Subtract the numbers:

16 - 4 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

n ab = n a n b

= 3 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 23

u_{1, 2} = \frac{- 4 \pm 2 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 4 + 2 3}{2 \cdot 1}, u_{2} = \frac{- 4 - 2 3}{2 \cdot 1}

u = \frac{- 4 + 2 3}{2 \cdot 1} : - 2 + 3

\frac{- 4 + 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 + 2 3}{2}

Factor

- 4 + 23 : 2 (- 2 + 3)

- 4 + 23

Rewrite as

= - 2 \cdot 2 + 23

Factor out common term

2

= 2 (- 2 + 3)

= \frac{2 ( - 2 + 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 2 + 3

u = \frac{- 4 - 2 3}{2 \cdot 1} : - 2 - 3

\frac{- 4 - 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4 - 2 3}{2}

Factor

- 4 - 23 : - 2 (2 + 3)

- 4 - 23

Rewrite as

= - 2 \cdot 2 - 23

Factor out common term

2

= - 2 (2 + 3)

= - \frac{2 ( 2 + 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (2 + 3)

Negate

- (2 + 3) = - 2 - 3

= - 2 - 3

The solutions to the quadratic equation are:

u = - 2 + 3, u = - 2 - 3

Substitute back

u = tan (x)

tan (x) = - 2 + 3, tan (x) = - 2 - 3

tan (x) = - 2 + 3, tan (x) = - 2 - 3

tan (x) = - 2 + 3 : x = arctan (- 2 + 3) + πn

tan (x) = - 2 + 3

Apply trig inverse properties

tan (x) = - 2 + 3

General solutions for

tan (x) = - 2 + 3

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

x = arctan (- 2 + 3) + πn

x = arctan (- 2 + 3) + πn

tan (x) = - 2 - 3 : x = arctan (- 2 - 3) + πn

tan (x) = - 2 - 3

Apply trig inverse properties

tan (x) = - 2 - 3

General solutions for

tan (x) = - 2 - 3

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

x = arctan (- 2 - 3) + πn

x = arctan (- 2 - 3) + πn

Combine all the solutions

x = arctan (- 2 + 3) + πn, x = arctan (- 2 - 3) + πn

Show solutions in decimal form

x = - 0.26179 \dots + πn, x = - 1.30899 \dots + πn

Graph

Popular Examples

2cosh^2(x)-5sinh(x)=5

2 cosh^{2} (x) - 5 sinh (x) = 5

10+5cos(2x)=15cos(x)

10 + 5 cos (2 x) = 15 cos (x)

solvefor y,x=cos(2y)

so l v e f or y, x = cos (2 y)

tan(x)= a/b

tan (x) = \frac{a}{b}

solvefor x,arcsin(3x-4y)= pi/2

so l v e f or x, arcsin (3 x - 4 y) = \frac{π}{2}

Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+4tan(x)+1=0 ?
The general solution for tan^2(x)+4tan(x)+1=0 is x=-0.26179…+pin,x=-1.30899…+pin