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

The general solution for 2tan^2(θ)+5tan(θ)=0 is θ=pin,θ=-1.19028…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2tan^2(θ)+5tan(θ)=0

Solution

2 tan^{2} (θ) + 5 tan (θ) = 0

Solution

θ = πn, θ = - 1.19028 \dots + πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = - 68.19859 \dots^{\circ} + 18 0^{\circ} n

Solution steps

2 tan^{2} (θ) + 5 tan (θ) = 0

Solve by substitution

2 tan^{2} (θ) + 5 tan (θ) = 0

Let:

tan (θ) = u

2 u^{2} + 5 u = 0

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

2 u^{2} + 5 u = 0

Solve with the quadratic formula

2 u^{2} + 5 u = 0

Quadratic Equation Formula:

For

a = 2, b = 5, c = 0

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 0 = 5

5^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

Apply radical rule:

assuming

a \geq 0

= 5

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

Separate the solutions

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

u = \frac{- 5 + 5}{2 \cdot 2} : 0

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

Add/Subtract the numbers:

- 5 + 5 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

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

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

Subtract the numbers:

- 5 - 5 = - 10

= \frac{- 10}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 10}{4}

Apply the fraction rule:

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

= - \frac{10}{4}

Cancel the common factor:

2

= - \frac{5}{2}

The solutions to the quadratic equation are:

u = 0, u = - \frac{5}{2}

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - \frac{5}{2}

tan (θ) = 0, tan (θ) = - \frac{5}{2}

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 0

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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

tan (θ) = - \frac{5}{2} : θ = arctan (- \frac{5}{2}) + πn

tan (θ) = - \frac{5}{2}

Apply trig inverse properties

tan (θ) = - \frac{5}{2}

General solutions for

tan (θ) = - \frac{5}{2}

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

θ = arctan (- \frac{5}{2}) + πn

θ = arctan (- \frac{5}{2}) + πn

Combine all the solutions

θ = πn, θ = arctan (- \frac{5}{2}) + πn

Show solutions in decimal form

θ = πn, θ = - 1.19028 \dots + πn

Popular Examples

cos(θ)-sqrt(3)sin(θ)=1

cos (θ) - 3 sin (θ) = 1

sec(x)= 1/2

sec (x) = \frac{1}{2}

2cos^2(x)=sin(x)+1

2 cos^{2} (x) = sin (x) + 1

csch(x)= 5/12

csch (x) = \frac{5}{12}

2sin^2(x)+sin(x)=1,0<= x<= 2pi

2 sin^{2} (x) + sin (x) = 1, 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 2tan^2(θ)+5tan(θ)=0 ?
The general solution for 2tan^2(θ)+5tan(θ)=0 is θ=pin,θ=-1.19028…+pin