What is the general solution for tan(2x)tan(x)=1 ?

The general solution for tan(2x)tan(x)=1 is x=0.52359…+pin,x=-0.52359…+pin

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tan(2x)tan(x)=1

Solution

tan (2 x) tan (x) = 1

Solution

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

Degrees

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

Solution steps

tan (2 x) tan (x) = 1

Subtract

1

from both sides

tan (2 x) tan (x) - 1 = 0

Rewrite using trig identities

- 1 + tan (2 x) tan (x)

Use the Double Angle identity:

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

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

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

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

Multiply fractions:

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

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

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

2 tan (x) tan (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

= 2 tan^{2} (x)

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

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

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

Multiply both sides by

1 - u^{2}

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

Multiply both sides by

1 - u^{2}

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

Simplify

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

Simplify

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

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

Multiply:

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

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

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

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

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

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

Solve

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

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

Expand

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

- (1 - u^{2}) + 2 u^{2}

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

- (1 - u^{2})

Distribute parentheses

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

Apply minus-plus rules

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

= - 1 + u^{2}

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

Add similar elements:

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

= - 1 + 3 u^{2}

- 1 + 3 u^{2} = 0

Move

1

to the right side

- 1 + 3 u^{2} = 0

Add

1

to both sides

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

Simplify

3 u^{2} = 1

3 u^{2} = 1

Divide both sides by

3

3 u^{2} = 1

Divide both sides by

3

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{3}, u = - \frac{1}{3}

u = \frac{1}{3}, u = - \frac{1}{3}

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

- 1 + \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 = \frac{1}{3}, u = - \frac{1}{3}

Substitute back

u = tan (x)

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

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

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

Apply trig inverse properties

tan (x) = \frac{1}{3}

General solutions for

tan (x) = \frac{1}{3}

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

Combine all the solutions

x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + πn

Show solutions in decimal form

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2x)tan(x)=1 ?
The general solution for tan(2x)tan(x)=1 is x=0.52359…+pin,x=-0.52359…+pin