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

The general solution for tan(x)=tan(2x) is x=pin

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tan(x)=tan(2x)

Solution

tan (x) = tan (2 x)

Solution

x = πn

Degrees

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

Solution steps

tan (x) = tan (2 x)

Subtract

tan (2 x)

from both sides

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

Rewrite using trig identities

- tan (2 x) + tan (x)

Use the Double Angle identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

- 2 tan (x) + tan (x) (1 - tan^{2} (x)) : - tan (x) - tan^{3} (x)

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

Expand

tan (x) (1 - tan^{2} (x)) : tan (x) - tan^{3} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = tan (x), b = 1, c = tan^{2} (x)

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

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

Simplify

1 \cdot tan (x) - tan^{2} (x) tan (x) : tan (x) - tan^{3} (x)

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

1 \cdot tan (x) = tan (x)

1 \cdot tan (x)

Multiply:

1 \cdot tan (x) = tan (x)

= tan (x)

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

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

Apply exponent rule:

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

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

= tan^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= tan^{3} (x)

= tan (x) - tan^{3} (x)

= tan (x) - tan^{3} (x)

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

Add similar elements:

- 2 tan (x) + tan (x) = - tan (x)

= - tan (x) - tan^{3} (x)

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

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

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- u - u^{3} = 0

Solve

- u - u^{3} = 0 : u = 0, u = i, u = - i

- u - u^{3} = 0

Factor

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

- u - u^{3}

Apply exponent rule:

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

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

= - u^{2} u - u

Factor out common term

- u

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u^{2} + 1 = 0

Solve

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

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = 0, u = i, u = - i

u = 0, u = i, u = - i

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

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

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = i, tan (x) = - i

tan (x) = 0, tan (x) = i, tan (x) = - i

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) = i :

No Solution

tan (x) = i

No Solution

tan (x) = - i :

No Solution

tan (x) = - i

No Solution

Combine all the solutions

x = πn

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Frequently Asked Questions (FAQ)

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