What is the general solution for 3tan(θ)=tan(2θ) ?

The general solution for 3tan(θ)=tan(2θ) is θ=pin,θ=(5pi)/6+pin,θ= pi/6+pin

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Popular Trigonometry >

3tan(θ)=tan(2θ)

Solution

3 tan (θ) = tan (2 θ)

Solution

θ = πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = 15 0^{\circ} + 18 0^{\circ} n, θ = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

3 tan (θ) = tan (2 θ)

Subtract

tan (2 θ)

from both sides

3 tan (θ) - tan (2 θ) = 0

Rewrite using trig identities

- tan (2 θ) + 3 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 )} + 3 tan (θ)

Simplify

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

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

Convert element to fraction:

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

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

Expand

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

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

Expand

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

3 tan (θ) (1 - tan^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3 tan (θ), b = 1, c = tan^{2} (θ)

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

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

Simplify

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

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

3 \cdot 1 \cdot tan (θ) = 3 tan (θ)

3 \cdot 1 \cdot tan (θ)

Multiply the numbers:

3 \cdot 1 = 3

= 3 tan (θ)

3 tan^{2} (θ) tan (θ) = 3 tan^{3} (θ)

3 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)

= 3 tan^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= 3 tan^{3} (θ)

= 3 tan (θ) - 3 tan^{3} (θ)

= 3 tan (θ) - 3 tan^{3} (θ)

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

Add similar elements:

- 2 tan (θ) + 3 tan (θ) = tan (θ)

= tan (θ) - 3 tan^{3} (θ)

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

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

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

Solve by substitution

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

Let:

tan (θ) = u

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

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

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

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

u - 3 u^{3} = 0

Solve

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

u - 3 u^{3} = 0

Factor

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

u - 3 u^{3}

Factor out common term

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

- 3 u^{3} + u

Apply exponent rule:

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

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

= - 3 u^{2} u + u

Factor out common term

- u

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

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

Factor

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

3 u^{2} - 1

Rewrite

3 u^{2} - 1

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

3 u^{2} - 1

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

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

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

= (3 u + 1) (3 u - 1)

= - u (3 u + 1) (3 u - 1)

- u (3 u + 1) (3 u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 3 u + 1 = 0 or 3 u - 1 = 0

Solve

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

3 u + 1 = 0

Move

1

to the right side

3 u + 1 = 0

Subtract

1

from both sides

3 u + 1 - 1 = 0 - 1

Simplify

3 u = - 1

3 u = - 1

Divide both sides by

3

3 u = - 1

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 u}{3} : u

\frac{3 u}{3}

Cancel the common factor:

3

= u

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

Solve

3 u - 1 = 0 : u = \frac{3}{3}

3 u - 1 = 0

Move

1

to the right side

3 u - 1 = 0

Add

1

to both sides

3 u - 1 + 1 = 0 + 1

Simplify

3 u = 1

3 u = 1

Divide both sides by

3

3 u = 1

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 u}{3} : u

\frac{3 u}{3}

Cancel the common factor:

3

= u

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

The solutions are

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

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

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

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

Substitute back

u = tan (θ)

tan (θ) = 0, tan (θ) = - \frac{3}{3}, tan (θ) = \frac{3}{3}

tan (θ) = 0, tan (θ) = - \frac{3}{3}, tan (θ) = \frac{3}{3}

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) = - \frac{3}{3} : θ = \frac{5 π}{6} + πn

tan (θ) = - \frac{3}{3}

General solutions for

tan (θ) = - \frac{3}{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}

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

tan (θ) = \frac{3}{3} : θ = \frac{π}{6} + πn

tan (θ) = \frac{3}{3}

General solutions for

tan (θ) = \frac{3}{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}

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

Combine all the solutions

θ = πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3tan(θ)=tan(2θ) ?
The general solution for 3tan(θ)=tan(2θ) is θ=pin,θ=(5pi)/6+pin,θ= pi/6+pin