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

The general solution for tan(x/2)tan(x)=-3 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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tan(x/2)tan(x)=-3

Solution

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

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

- 3

from both sides

tan (\frac{x}{2}) tan (x) + 3 = 0

Let:

u = \frac{x}{2}

tan (u) tan (2 u) + 3 = 0

Rewrite using trig identities

3 + tan (2 u) tan (u)

Use the Double Angle identity:

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

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

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

\frac{2 tan ( u )}{1 - tan ^{2} ( u )} tan (u)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

2 tan (u) tan (u) = 2 tan^{2} (u)

2 tan (u) tan (u)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (u) tan (u) = tan^{1 + 1} (u)

= 2 tan^{1 + 1} (u)

Add the numbers:

1 + 1 = 2

= 2 tan^{2} (u)

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

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

3 + \frac{2 tan ^{2} ( u )}{1 - tan ^{2} ( u )} = 0

Solve by substitution

3 + \frac{2 tan ^{2} ( u )}{1 - tan ^{2} ( u )} = 0

Let:

tan (u) = u

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

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

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

Multiply both sides by

1 - u^{2}

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

Multiply both sides by

1 - u^{2}

3 (1 - u^{2}) + \frac{2 u ^{2}}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

Simplify

3 (1 - u^{2}) + \frac{2 u ^{2}}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

Simplify

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

1 - u^{2}

= 2 u^{2}

Simplify

0 \cdot (1 - u^{2}) : 0

0 \cdot (1 - u^{2})

Apply rule

0 \cdot a = 0

= 0

3 (1 - u^{2}) + 2 u^{2} = 0

3 (1 - u^{2}) + 2 u^{2} = 0

3 (1 - u^{2}) + 2 u^{2} = 0

Solve

3 (1 - u^{2}) + 2 u^{2} = 0 : u = 3, u = - 3

3 (1 - u^{2}) + 2 u^{2} = 0

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}

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 u^{2}

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

Add similar elements:

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

3 - u^{2} = 0

Move

3

to the right side

3 - u^{2} = 0

Subtract

3

from both sides

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

Simplify

- u^{2} = - 3

- u^{2} = - 3

Divide both sides by

- 1

- u^{2} = - 3

Divide both sides by

- 1

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

Simplify

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

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

u = 3, u = - 3

u = 3, u = - 3

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

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

Substitute back

u = tan (u)

tan (u) = 3, tan (u) = - 3

tan (u) = 3, tan (u) = - 3

tan (u) = 3 : u = \frac{π}{3} + πn

tan (u) = 3

General solutions for

tan (u) = 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}

u = \frac{π}{3} + πn

u = \frac{π}{3} + πn

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

tan (u) = - 3

General solutions for

tan (u) = - 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}

u = \frac{2 π}{3} + πn

u = \frac{2 π}{3} + πn

Combine all the solutions

u = \frac{π}{3} + πn, u = \frac{2 π}{3} + πn

Substitute back

u = \frac{x}{2}

\frac{x}{2} = \frac{π}{3} + πn : x = \frac{2 π}{3} + 2 πn

\frac{x}{2} = \frac{π}{3} + πn

Multiply both sides by

2

\frac{x}{2} = \frac{π}{3} + πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{π}{3} + 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{π}{3} + 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 \cdot \frac{π}{3} + 2 πn : \frac{2 π}{3} + 2 πn

2 \cdot \frac{π}{3} + 2 πn

Multiply

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

\frac{x}{2} = \frac{2 π}{3} + πn : x = \frac{4 π}{3} + 2 πn

\frac{x}{2} = \frac{2 π}{3} + πn

Multiply both sides by

2

\frac{x}{2} = \frac{2 π}{3} + πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{2 π}{3} + 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{2 π}{3} + 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 \cdot \frac{2 π}{3} + 2 πn : \frac{4 π}{3} + 2 πn

2 \cdot \frac{2 π}{3} + 2 πn

2 \cdot \frac{2 π}{3} = \frac{4 π}{3}

2 \cdot \frac{2 π}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 π 2}{3}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 π}{3}

= \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

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