What is the general solution for 2tan^2(x)+11tan(x)+9=0 ?

The general solution for 2tan^2(x)+11tan(x)+9=0 is x=(3pi)/4+pin,x=-1.35212…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2tan^2(x)+11tan(x)+9=0

Solution

2 tan^{2} (x) + 11 tan (x) + 9 = 0

Solution

x = \frac{3 π}{4} + πn, x = - 1.35212 \dots + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n, x = - 77.47119 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 tan^{2} (x) + 11 tan (x) + 9 = 0

Solve by substitution

2 tan^{2} (x) + 11 tan (x) + 9 = 0

Let:

tan (x) = u

2 u^{2} + 11 u + 9 = 0

2 u^{2} + 11 u + 9 = 0 : u = - 1, u = - \frac{9}{2}

2 u^{2} + 11 u + 9 = 0

Solve with the quadratic formula

2 u^{2} + 11 u + 9 = 0

Quadratic Equation Formula:

For

a = 2, b = 11, c = 9

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 \cdot 9}{2 \cdot 2}

1 1^{2} - 4 \cdot 2 \cdot 9 = 7

1 1^{2} - 4 \cdot 2 \cdot 9

Multiply the numbers:

4 \cdot 2 \cdot 9 = 72

= 1 1^{2} - 72

1 1^{2} = 121

= 121 - 72

Subtract the numbers:

121 - 72 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 11 \pm 7}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 11 + 7}{2 \cdot 2}, u_{2} = \frac{- 11 - 7}{2 \cdot 2}

u = \frac{- 11 + 7}{2 \cdot 2} : - 1

\frac{- 11 + 7}{2 \cdot 2}

Add/Subtract the numbers:

- 11 + 7 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 11 - 7}{2 \cdot 2} : - \frac{9}{2}

\frac{- 11 - 7}{2 \cdot 2}

Subtract the numbers:

- 11 - 7 = - 18

= \frac{- 18}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 18}{4}

Apply the fraction rule:

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

= - \frac{18}{4}

Cancel the common factor:

2

= - \frac{9}{2}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{9}{2}

Substitute back

u = tan (x)

tan (x) = - 1, tan (x) = - \frac{9}{2}

tan (x) = - 1, tan (x) = - \frac{9}{2}

tan (x) = - 1 : x = \frac{3 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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

tan (x) = - \frac{9}{2} : x = arctan (- \frac{9}{2}) + πn

tan (x) = - \frac{9}{2}

Apply trig inverse properties

tan (x) = - \frac{9}{2}

General solutions for

tan (x) = - \frac{9}{2}

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

x = arctan (- \frac{9}{2}) + πn

x = arctan (- \frac{9}{2}) + πn

Combine all the solutions

x = \frac{3 π}{4} + πn, x = arctan (- \frac{9}{2}) + πn

Show solutions in decimal form

x = \frac{3 π}{4} + πn, x = - 1.35212 \dots + πn

Popular Examples

cos(x)=0.37

cos (x) = 0.37

2cos(2θ)-sqrt(3)=0

2 cos (2 θ) - 3 = 0

3arccos(4x)=2pi

3 arccos (4 x) = 2 π

cos(3x)=cos(x)

cos (3 x) = cos (x)

tan(x+32)=cot(x-20)

tan (x + 3 2^{\circ}) = cot (x - 2 0^{\circ})

Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(x)+11tan(x)+9=0 ?
The general solution for 2tan^2(x)+11tan(x)+9=0 is x=(3pi)/4+pin,x=-1.35212…+pin