What is the general solution for (2tan(x))/(1-tan^2(x))=sqrt(3) ?

The general solution for (2tan(x))/(1-tan^2(x))=sqrt(3) is x=(2pi)/3+pin,x= pi/6+pin

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(2tan(x))/(1-tan^2(x))=sqrt(3)

Solution

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

Solution

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

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Multiply both sides by

1 - u^{2}

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

Multiply both sides by

1 - u^{2}

\frac{2 u}{1 - u ^{2}} (1 - u^{2}) = 3 (1 - u^{2})

Simplify

2 u = 3 (1 - u^{2})

2 u = 3 (1 - u^{2})

Solve

2 u = 3 (1 - u^{2}) : u = - 3, u = \frac{3}{3}

2 u = 3 (1 - u^{2})

Expand

3 (1 - u^{2}) : 3 - 3 u^{2}

3 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = u^{2}

= 3 \cdot 1 - 3 u^{2}

= 1 \cdot 3 - 3 u^{2}

Multiply:

1 \cdot 3 = 3

= 3 - 3 u^{2}

2 u = 3 - 3 u^{2}

Switch sides

3 - 3 u^{2} = 2 u

Move

2 u

to the left side

3 - 3 u^{2} = 2 u

Subtract

2 u

from both sides

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

Simplify

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

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

Write in the standard form

a x^{2} + b x + c = 0

- 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} - 4 (- 3) 3 = 4

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

Apply rule

- (- a) = a

= (- 2)^{2} + 433

(- 2)^{2} = 2^{2}

(- 2)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2}

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 )} : - 3

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Add 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

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Subtract the numbers:

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

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

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

\frac{2 u}{1 - u ^{2}}

and compare to zero

Solve

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

Move

1

to the right side

1 - u^{2} = 0

Subtract

1

from both sides

1 - u^{2} - 1 = 0 - 1

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The following points are undefined

u = 1, u = - 1

Combine undefined points with solutions:

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

Substitute back

u = tan (x)

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

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

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

tan (x) = - 3

General solutions for

tan (x) = - 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{2 π}{3} + πn

x = \frac{2 π}{3} + πn

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

x = \frac{π}{6} + πn

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for (2tan(x))/(1-tan^2(x))=sqrt(3) ?
The general solution for (2tan(x))/(1-tan^2(x))=sqrt(3) is x=(2pi)/3+pin,x= pi/6+pin