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

The general solution for tan^3(x)=2 is x=0.89990…+pin

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Popular Trigonometry >

tan^3(x)=2

Solution

tan^{3} (x) = 2

Solution

x = 0.89990 \dots + πn

Degrees

x = 51.56095 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan^{3} (x) = 2

Solve by substitution

tan^{3} (x) = 2

Let:

tan (x) = u

u^{3} = 2

u^{3} = 2 : u = 32, u = - \frac{3 2}{2} + i \frac{3 2 3}{2}, u = - \frac{3 2}{2} - i \frac{3 2 3}{2}

u^{3} = 2

For

x^{3} = f (a)

the solutions are

x = 3 f (a), 3 f (a) \frac{- 1 - 3 i}{2}, 3 f (a) \frac{- 1 + 3 i}{2}

u = 32, u = 32 \frac{- 1 + 3 i}{2}, u = 32 \frac{- 1 - 3 i}{2}

Simplify

32 \frac{- 1 + 3 i}{2} : - \frac{3 2}{2} + i \frac{3 2 3}{2}

32 \frac{- 1 + 3 i}{2}

Multiply fractions:

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

= \frac{( - 1 + 3 i ) 3 2}{2}

Apply radical rule:

n a = a^{\frac{1}{n}}

32 = 2^{\frac{1}{3}}

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 + 3 i}{2 ^{1 - \frac{1}{3}}}

Subtract the numbers:

1 - \frac{1}{3} = \frac{2}{3}

= \frac{- 1 + 3 i}{2 ^{\frac{2}{3}}}

Rationalize

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

\frac{- 1 + 3 i}{2 ^{\frac{2}{3}}}

Multiply by the conjugate

\frac{3 2}{3 2}

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

2^{\frac{2}{3}} 32 = 2

2^{\frac{2}{3}} 32

Apply exponent rule:

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

2^{\frac{2}{3}} 32 = 2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} = 2^{\frac{2}{3} + \frac{1}{3}}

= 2^{\frac{2}{3} + \frac{1}{3}}

Join

\frac{2}{3} + \frac{1}{3} : 1

\frac{2}{3} + \frac{1}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 1}{3}

Add the numbers:

2 + 1 = 3

= \frac{3}{3}

Apply rule

\frac{a}{a} = 1

= 1

= 2^{1}

Apply rule

a^{1} = a

= 2

= \frac{3 2 ( - 1 + 3 i )}{2}

= \frac{3 2 ( - 1 + 3 i )}{2}

Rewrite

\frac{3 2 ( - 1 + 3 i )}{2}

in standard complex form:

- \frac{3 2}{2} + \frac{3 3 2}{2} i

\frac{3 2 ( - 1 + 3 i )}{2}

Apply radical rule:

n a = a^{\frac{1}{n}}

32 = 2^{\frac{1}{3}}

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 + 3 i}{2 ^{1 - \frac{1}{3}}}

Subtract the numbers:

1 - \frac{1}{3} = \frac{2}{3}

= \frac{- 1 + 3 i}{2 ^{\frac{2}{3}}}

Apply the fraction rule:

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

\frac{- 1 + 3 i}{2 ^{\frac{2}{3}}} = - \frac{1}{2 ^{\frac{2}{3}}} + \frac{3 i}{2 ^{\frac{2}{3}}}

= - \frac{1}{2 ^{\frac{2}{3}}} + \frac{3 i}{2 ^{\frac{2}{3}}}

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

\frac{3}{2 ^{\frac{2}{3}}}

Multiply by the conjugate

\frac{3 2}{3 2}

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

2^{\frac{2}{3}} 32 = 2

2^{\frac{2}{3}} 32

Apply exponent rule:

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

2^{\frac{2}{3}} 32 = 2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} = 2^{\frac{2}{3} + \frac{1}{3}}

= 2^{\frac{2}{3} + \frac{1}{3}}

Join

\frac{2}{3} + \frac{1}{3} : 1

\frac{2}{3} + \frac{1}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 1}{3}

Add the numbers:

2 + 1 = 3

= \frac{3}{3}

Apply rule

\frac{a}{a} = 1

= 1

= 2^{1}

Apply rule

a^{1} = a

= 2

= \frac{3 3 2}{2}

= - \frac{1}{2 ^{\frac{2}{3}}} + \frac{3 3 2}{2} i

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

- \frac{1}{2 ^{\frac{2}{3}}}

Multiply by the conjugate

\frac{3 2}{3 2}

= - \frac{1 \cdot 3 2}{2 ^{\frac{2}{3}} 3 2}

1 \cdot 32 = 32

2^{\frac{2}{3}} 32 = 2

2^{\frac{2}{3}} 32

Apply exponent rule:

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

2^{\frac{2}{3}} 32 = 2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} = 2^{\frac{2}{3} + \frac{1}{3}}

= 2^{\frac{2}{3} + \frac{1}{3}}

Join

\frac{2}{3} + \frac{1}{3} : 1

\frac{2}{3} + \frac{1}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 1}{3}

Add the numbers:

2 + 1 = 3

= \frac{3}{3}

Apply rule

\frac{a}{a} = 1

= 1

= 2^{1}

Apply rule

a^{1} = a

= 2

= - \frac{3 2}{2}

= - \frac{3 2}{2} + \frac{3 3 2}{2} i

= - \frac{3 2}{2} + \frac{3 3 2}{2} i

Simplify

32 \frac{- 1 - 3 i}{2} : - \frac{3 2}{2} - i \frac{3 2 3}{2}

32 \frac{- 1 - 3 i}{2}

Multiply fractions:

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

= \frac{( - 1 - 3 i ) 3 2}{2}

Apply radical rule:

n a = a^{\frac{1}{n}}

32 = 2^{\frac{1}{3}}

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 - 3 i}{2 ^{1 - \frac{1}{3}}}

Subtract the numbers:

1 - \frac{1}{3} = \frac{2}{3}

= \frac{- 1 - 3 i}{2 ^{\frac{2}{3}}}

Rationalize

\frac{- 1 - 3 i}{2 ^{\frac{2}{3}}} : \frac{3 2 ( - 1 - 3 i )}{2}

\frac{- 1 - 3 i}{2 ^{\frac{2}{3}}}

Multiply by the conjugate

\frac{3 2}{3 2}

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

2^{\frac{2}{3}} 32 = 2

2^{\frac{2}{3}} 32

Apply exponent rule:

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

2^{\frac{2}{3}} 32 = 2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} = 2^{\frac{2}{3} + \frac{1}{3}}

= 2^{\frac{2}{3} + \frac{1}{3}}

Join

\frac{2}{3} + \frac{1}{3} : 1

\frac{2}{3} + \frac{1}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 1}{3}

Add the numbers:

2 + 1 = 3

= \frac{3}{3}

Apply rule

\frac{a}{a} = 1

= 1

= 2^{1}

Apply rule

a^{1} = a

= 2

= \frac{3 2 ( - 1 - 3 i )}{2}

= \frac{3 2 ( - 1 - 3 i )}{2}

Rewrite

\frac{3 2 ( - 1 - 3 i )}{2}

in standard complex form:

- \frac{3 2}{2} - \frac{3 3 2}{2} i

\frac{3 2 ( - 1 - 3 i )}{2}

Apply radical rule:

n a = a^{\frac{1}{n}}

32 = 2^{\frac{1}{3}}

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{- 1 - 3 i}{2 ^{1 - \frac{1}{3}}}

Subtract the numbers:

1 - \frac{1}{3} = \frac{2}{3}

= \frac{- 1 - 3 i}{2 ^{\frac{2}{3}}}

Apply the fraction rule:

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

\frac{- 1 - 3 i}{2 ^{\frac{2}{3}}} = - \frac{1}{2 ^{\frac{2}{3}}} - \frac{3 i}{2 ^{\frac{2}{3}}}

= - \frac{1}{2 ^{\frac{2}{3}}} - \frac{3 i}{2 ^{\frac{2}{3}}}

- \frac{3}{2 ^{\frac{2}{3}}} = - \frac{3 3 2}{2}

- \frac{3}{2 ^{\frac{2}{3}}}

Multiply by the conjugate

\frac{3 2}{3 2}

= - \frac{3 3 2}{2 ^{\frac{2}{3}} 3 2}

2^{\frac{2}{3}} 32 = 2

2^{\frac{2}{3}} 32

Apply exponent rule:

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

2^{\frac{2}{3}} 32 = 2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} = 2^{\frac{2}{3} + \frac{1}{3}}

= 2^{\frac{2}{3} + \frac{1}{3}}

Join

\frac{2}{3} + \frac{1}{3} : 1

\frac{2}{3} + \frac{1}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 1}{3}

Add the numbers:

2 + 1 = 3

= \frac{3}{3}

Apply rule

\frac{a}{a} = 1

= 1

= 2^{1}

Apply rule

a^{1} = a

= 2

= - \frac{3 3 2}{2}

= - \frac{1}{2 ^{\frac{2}{3}}} - \frac{3 3 2}{2} i

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

- \frac{1}{2 ^{\frac{2}{3}}}

Multiply by the conjugate

\frac{3 2}{3 2}

= - \frac{1 \cdot 3 2}{2 ^{\frac{2}{3}} 3 2}

1 \cdot 32 = 32

2^{\frac{2}{3}} 32 = 2

2^{\frac{2}{3}} 32

Apply exponent rule:

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

2^{\frac{2}{3}} 32 = 2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} = 2^{\frac{2}{3} + \frac{1}{3}}

= 2^{\frac{2}{3} + \frac{1}{3}}

Join

\frac{2}{3} + \frac{1}{3} : 1

\frac{2}{3} + \frac{1}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 1}{3}

Add the numbers:

2 + 1 = 3

= \frac{3}{3}

Apply rule

\frac{a}{a} = 1

= 1

= 2^{1}

Apply rule

a^{1} = a

= 2

= - \frac{3 2}{2}

= - \frac{3 2}{2} - \frac{3 3 2}{2} i

= - \frac{3 2}{2} - \frac{3 3 2}{2} i

u = 32, u = - \frac{3 2}{2} + i \frac{3 2 3}{2}, u = - \frac{3 2}{2} - i \frac{3 2 3}{2}

Substitute back

u = tan (x)

tan (x) = 32, tan (x) = - \frac{3 2}{2} + i \frac{3 2 3}{2}, tan (x) = - \frac{3 2}{2} - i \frac{3 2 3}{2}

tan (x) = 32, tan (x) = - \frac{3 2}{2} + i \frac{3 2 3}{2}, tan (x) = - \frac{3 2}{2} - i \frac{3 2 3}{2}

tan (x) = 32 : x = arctan (32) + πn

tan (x) = 32

Apply trig inverse properties

tan (x) = 32

General solutions for

tan (x) = 32

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (32) + πn

x = arctan (32) + πn

tan (x) = - \frac{3 2}{2} + i \frac{3 2 3}{2} :

No Solution

tan (x) = - \frac{3 2}{2} + i \frac{3 2 3}{2}

No Solution

tan (x) = - \frac{3 2}{2} - i \frac{3 2 3}{2} :

No Solution

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

No Solution

Combine all the solutions

x = arctan (32) + πn

Show solutions in decimal form

x = 0.89990 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan^3(x)=2 ?
The general solution for tan^3(x)=2 is x=0.89990…+pin