What is the general solution for 2tan^2(x)+5tan(x)+3=0 ?

The general solution for 2tan^2(x)+5tan(x)+3=0 is x=(3pi)/4+pin,x=-0.98279…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2tan^2(x)+5tan(x)+3=0

Solution

2 tan^{2} (x) + 5 tan (x) + 3 = 0

Solution

x = \frac{3 π}{4} + πn, x = - 0.98279 \dots + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n, x = - 56.30993 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 tan^{2} (x) + 5 tan (x) + 3 = 0

Solve by substitution

2 tan^{2} (x) + 5 tan (x) + 3 = 0

Let:

tan (x) = u

2 u^{2} + 5 u + 3 = 0

2 u^{2} + 5 u + 3 = 0 : u = - 1, u = - \frac{3}{2}

2 u^{2} + 5 u + 3 = 0

Solve with the quadratic formula

2 u^{2} + 5 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = 5, c = 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 3 = 1

5^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 5 + 1 = - 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{- 5 - 1}{2 \cdot 2} : - \frac{3}{2}

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

Subtract the numbers:

- 5 - 1 = - 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{4}

Apply the fraction rule:

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

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{3}{2}

Substitute back

u = tan (x)

tan (x) = - 1, tan (x) = - \frac{3}{2}

tan (x) = - 1, tan (x) = - \frac{3}{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{3}{2} : x = arctan (- \frac{3}{2}) + πn

tan (x) = - \frac{3}{2}

Apply trig inverse properties

tan (x) = - \frac{3}{2}

General solutions for

tan (x) = - \frac{3}{2}

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

x = arctan (- \frac{3}{2}) + πn

x = arctan (- \frac{3}{2}) + πn

Combine all the solutions

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

Show solutions in decimal form

x = \frac{3 π}{4} + πn, x = - 0.98279 \dots + πn

Popular Examples

csc^2(x)=4

csc^{2} (x) = 4

3tanh(2θ)=5sech(θ)+1

3 tanh (2 θ) = 5 sech (θ) + 1

2cos^2(θ)+cos(θ)=1

2 cos^{2} (θ) + cos (θ) = 1

tan(x)sin(x)-sin(x)=0

tan (x) sin (x) - sin (x) = 0

2sin(x)=csc(x)

2 sin (x) = csc (x)

Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(x)+5tan(x)+3=0 ?
The general solution for 2tan^2(x)+5tan(x)+3=0 is x=(3pi)/4+pin,x=-0.98279…+pin