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

The general solution for sqrt(3)tan^2(x)+tan(x)=0 is x=pin,x=(5pi)/6+pin

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Popular >

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

Solution

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

Solution

x = πn, x = \frac{5 π}{6} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

3 u^{2} + u = 0

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

3 u^{2} + u = 0

Solve with the quadratic formula

3 u^{2} + u = 0

Quadratic Equation Formula:

For

a = 3, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 3 \cdot 0}{2 3}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 3 \cdot 0}{2 3}

1^{2} - 43 \cdot 0 = 1

1^{2} - 43 \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 0 \cdot 3

Apply rule

0 \cdot a = 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 3}

Apply rule

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

= 0

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

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

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{2 3}

Apply the fraction rule:

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

= - \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}

The solutions to the quadratic equation are:

u = 0, u = - \frac{3}{3}

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = - \frac{3}{3}

tan (x) = 0, tan (x) = - \frac{3}{3}

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 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}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = - \frac{3}{3} : x = \frac{5 π}{6} + πn

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

General solutions for

tan (x) = - \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}

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

Combine all the solutions

x = πn, x = \frac{5 π}{6} + πn

Popular Examples

tan^2(x)-4sin(x)+4=0

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

1/(sin(x))+1/(cos(x))=0

\frac{1}{sin ( x )} + \frac{1}{cos ( x )} = 0

1-cos(2x)=0

1 - cos (2 x) = 0

sin(x)= 37/64

sin (x) = \frac{37}{64}

1=4tan(-4/81+c)

1 = 4 tan (- \frac{4}{81} + c)

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)tan^2(x)+tan(x)=0 ?
The general solution for sqrt(3)tan^2(x)+tan(x)=0 is x=pin,x=(5pi)/6+pin