What is the general solution for tan(2θ)-tan(θ)=0 ?

The general solution for tan(2θ)-tan(θ)=0 is θ=pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan(2θ)-tan(θ)=0

Solution

tan (2 θ) - tan (θ) = 0

Solution

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

tan (2 θ) - tan (θ) = 0

Rewrite using trig identities

tan (2 θ) - tan (θ)

Use the Double Angle identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

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

2 tan (θ) - tan (θ) (1 - tan^{2} (θ))

Expand

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

- tan (θ) (1 - tan^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - tan (θ), b = 1, c = tan^{2} (θ)

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

Apply minus-plus rules

- (- a) = a

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

Simplify

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

- 1 \cdot tan (θ) + tan^{2} (θ) tan (θ)

1 \cdot tan (θ) = tan (θ)

1 \cdot tan (θ)

Multiply:

1 \cdot tan (θ) = tan (θ)

= tan (θ)

tan^{2} (θ) tan (θ) = tan^{3} (θ)

tan^{2} (θ) tan (θ)

Apply exponent rule:

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

tan^{2} (θ) tan (θ) = tan^{2 + 1} (θ)

= tan^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= tan^{3} (θ)

= - tan (θ) + tan^{3} (θ)

= - tan (θ) + tan^{3} (θ)

= 2 tan (θ) - tan (θ) + tan^{3} (θ)

Add similar elements:

2 tan (θ) - tan (θ) = tan (θ)

= tan (θ) + tan^{3} (θ)

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

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

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

Solve by substitution

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

Let:

tan (θ) = 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 (θ)

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

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

tan (θ) = 0 : θ = πn

tan (θ) = 0

General solutions for

tan (θ) = 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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

tan (θ) = i :

No Solution

tan (θ) = i

No Solution

tan (θ) = - i :

No Solution

tan (θ) = - i

No Solution

Combine all the solutions

θ = πn

Popular Examples

cos(t)=-1/2

cos (t) = - \frac{1}{2}

cos(2θ)+cos(θ)=0

cos (2 θ) + cos (θ) = 0

4sin(x)=2

4 sin (x) = 2

tan(4x)=1

tan (4 x) = 1

cos(3x)+cos(x)=0

cos (3 x) + cos (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan(2θ)-tan(θ)=0 ?
The general solution for tan(2θ)-tan(θ)=0 is θ=pin