What is the general solution for tan(x)+tan(2x)=0 ?

The general solution for tan(x)+tan(2x)=0 is x=pin,x=(2pi)/3+pin,x= pi/3+pin

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tan(x)+tan(2x)=0

Solution

tan (x) + tan (2 x) = 0

Solution

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

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

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{3 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)) : 3 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) = 3 tan (x)

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

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

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

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

Solve by substitution

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

Let:

tan (x) = u

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

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

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

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

- u^{3} + 3 u = 0

Solve

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

- u^{3} + 3 u = 0

Factor

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

- u^{3} + 3 u

Factor out common term

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

- u^{3} + 3 u

Apply exponent rule:

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

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

= - u^{2} u + 3 u

Factor out common term

- u

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

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

Factor

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

u^{2} - 3

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (u + 3) (u - 3)

= - u (u + 3) (u - 3)

- u (u + 3) (u - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or u + 3 = 0 or u - 3 = 0

Solve

u + 3 = 0 : u = - 3

u + 3 = 0

Move

3

to the right side

u + 3 = 0

Subtract

3

from both sides

u + 3 - 3 = 0 - 3

Simplify

u = - 3

u = - 3

Solve

u - 3 = 0 : u = 3

u - 3 = 0

Move

3

to the right side

u - 3 = 0

Add

3

to both sides

u - 3 + 3 = 0 + 3

Simplify

u = 3

u = 3

The solutions are

u = 0, u = - 3, u = 3

u = 0, u = - 3, u = 3

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

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

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = - 3, tan (x) = 3

tan (x) = 0, tan (x) = - 3, tan (x) = 3

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) = - 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) = 3 : x = \frac{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{π}{3} + πn

x = \frac{π}{3} + πn

Combine all the solutions

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

Popular Examples

3cos(2θ)+2=-cos(θ)

3 cos (2 θ) + 2 = - cos (θ)

cos(x)=(7.2)/(9.4)

cos (x) = \frac{7.2}{9.4}

sin(θ)=-sqrt(2)-cos(θ)

sin (θ) = - 2 - cos (θ)

sin(2θ)=3cos(2θ)

sin (2 θ) = 3 cos (2 θ)

1-cos(2x)=sin(2x)

1 - cos (2 x) = sin (2 x)

Frequently Asked Questions (FAQ)

What is the general solution for tan(x)+tan(2x)=0 ?
The general solution for tan(x)+tan(2x)=0 is x=pin,x=(2pi)/3+pin,x= pi/3+pin