What is the general solution for sqrt(3)tan^2(θ)+2tan(θ)-sqrt(3)=0 ?

The general solution for sqrt(3)tan^2(θ)+2tan(θ)-sqrt(3)=0 is θ= pi/6+pin,θ=(2pi)/3+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(3)tan^2(θ)+2tan(θ)-sqrt(3)=0

Solution

3 tan^{2} (θ) + 2 tan (θ) - 3 = 0

Solution

θ = \frac{π}{6} + πn, θ = \frac{2 π}{3} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

3 tan^{2} (θ) + 2 tan (θ) - 3 = 0

Solve by substitution

3 tan^{2} (θ) + 2 tan (θ) - 3 = 0

Let:

tan (θ) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = 2, c = - 3

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 3 ( - 3 )}{2 3}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 3 ( - 3 )}{2 3}

2^{2} - 43 (- 3) = 4

2^{2} - 43 (- 3)

Apply rule

- (- a) = a

= 2^{2} + 433

433 = 12

433

Apply radical rule:

a a = a

33 = 3

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 2 + 4 = 2

= \frac{2}{2 3}

Divide the numbers:

\frac{2}{2} = 1

= \frac{1}{3}

Rationalize

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

= \frac{3}{3}

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

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

Subtract the numbers:

- 2 - 4 = - 6

= \frac{- 6}{2 3}

Apply the fraction rule:

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

= - \frac{6}{2 3}

Divide the numbers:

\frac{6}{2} = 3

= \frac{3}{3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{3}{3 ^{\frac{1}{2}}}

Apply exponent rule:

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

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= 3^{1 - \frac{1}{2}}

Subtract the numbers:

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

= 3^{\frac{1}{2}}

Apply radical rule:

3^{\frac{1}{2}} = 3

= - 3

The solutions to the quadratic equation are:

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

Substitute back

u = tan (θ)

tan (θ) = \frac{3}{3}, tan (θ) = - 3

tan (θ) = \frac{3}{3}, tan (θ) = - 3

tan (θ) = \frac{3}{3} : θ = \frac{π}{6} + πn

tan (θ) = \frac{3}{3}

General solutions for

tan (θ) = \frac{3}{3}

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 = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

tan (θ) = - 3 : θ = \frac{2 π}{3} + πn

tan (θ) = - 3

General solutions for

tan (θ) = - 3

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 = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

Combine all the solutions

θ = \frac{π}{6} + πn, θ = \frac{2 π}{3} + πn

Popular Examples

sinh(x)= 1/2

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

sin^2(x)*sec(x)=(sin^2(x))/(cot(x))

sin^{2} (x) \cdot sec (x) = \frac{sin ^{2} ( x )}{cot ( x )}

2tan(θ)+sin(θ)cos(θ)=0

2 tan (θ) + sin (θ) cos (θ) = 0

sec(x)= 25/7

sec (x) = \frac{25}{7}

sin(x/2)= 1/2 sin(x)

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

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)tan^2(θ)+2tan(θ)-sqrt(3)=0 ?
The general solution for sqrt(3)tan^2(θ)+2tan(θ)-sqrt(3)=0 is θ= pi/6+pin,θ=(2pi)/3+pin