What is the general solution for tan(2θ)+tan(2θ)=tan(4θ) ?

The general solution for tan(2θ)+tan(2θ)=tan(4θ) is θ=(pin)/2

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tan(2θ)+tan(2θ)=tan(4θ)

Solution

tan (2 θ) + tan (2 θ) = tan (4 θ)

Solution

θ = \frac{πn}{2}

Degrees

θ = 0^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 θ) + tan (2 θ) = tan (4 θ)

Subtract

tan (4 θ)

from both sides

2 tan (2 θ) - tan (4 θ) = 0

Let:

u = 2 θ

2 tan (u) - tan (2 u) = 0

Rewrite using trig identities

- tan (2 u) + 2 tan (u)

Use the Double Angle identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

2 \cdot 1 \cdot tan (u) = 2 tan (u)

2 \cdot 1 \cdot tan (u)

Multiply the numbers:

2 \cdot 1 = 2

= 2 tan (u)

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

2 tan^{2} (u) tan (u)

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 tan^{3} (u)

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

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

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

Add similar elements:

- 2 tan (u) + 2 tan (u) = 0

= - 2 tan^{3} (u)

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

Apply the fraction rule:

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

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

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

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

Solve by substitution

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

Let:

tan (u) = u

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

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

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

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

2 u^{3} = 0

Solve

2 u^{3} = 0 : u = 0

2 u^{3} = 0

Divide both sides by

2

2 u^{3} = 0

Divide both sides by

2

2 u^{3} = 0

Divide both sides by

2

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

Simplify

u^{3} = 0

u^{3} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

u = 0

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

- \frac{2 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

Substitute back

u = tan (u)

tan (u) = 0

tan (u) = 0

tan (u) = 0 : u = πn

tan (u) = 0

General solutions for

tan (u) = 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}

u = 0 + πn

u = 0 + πn

Solve

u = 0 + πn : u = πn

u = 0 + πn

0 + πn = πn

u = πn

u = πn

Combine all the solutions

u = πn

Substitute back

u = 2 θ

2 θ = πn : θ = \frac{πn}{2}

2 θ = πn

Divide both sides by

2

2 θ = πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{πn}{2}

Simplify

θ = \frac{πn}{2}

θ = \frac{πn}{2}

θ = \frac{πn}{2}

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tanh (x) = \frac{12}{13}

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2 sin^{2} (x) = 5 sin (x) - 3

Frequently Asked Questions (FAQ)

What is the general solution for tan(2θ)+tan(2θ)=tan(4θ) ?
The general solution for tan(2θ)+tan(2θ)=tan(4θ) is θ=(pin)/2