What is the general solution for arctan(x+1/3)+arctan(x-1/3)=arctan(2) ?

The general solution for arctan(x+1/3)+arctan(x-1/3)=arctan(2) is x= 2/3

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arctan(x+1/3)+arctan(x-1/3)=arctan(2)

Solution

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

Solution

x = \frac{2}{3}

Solution steps

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

Rewrite using trig identities

arctan (x + \frac{1}{3}) + arctan (x - \frac{1}{3})

Use the Sum to Product identity:

arctan (s) + arctan (t) = arctan (\frac{s + t}{1 - s t})

= arctan (\frac{x + \frac{1}{3} + x - \frac{1}{3}}{1 - ( x + \frac{1}{3} ) ( x - \frac{1}{3} )})

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

Apply trig inverse properties

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

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

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

tan (arctan (2)) = 2

tan (arctan (2))

Rewrite using trig identities:

tan (arctan (2)) = 2

Use the following identity:

tan (arctan (x)) = x

= 2

= 2

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

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

Solve

\frac{x + \frac{1}{3} + x - \frac{1}{3}}{1 - ( x + \frac{1}{3} ) ( x - \frac{1}{3} )} = 2 : x = - \frac{5}{3}, x = \frac{2}{3}

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

Simplify

\frac{x + \frac{1}{3} + x - \frac{1}{3}}{1 - ( x + \frac{1}{3} ) ( x - \frac{1}{3} )} : \frac{18 x}{- 9 x ^{2} + 10}

\frac{x + \frac{1}{3} + x - \frac{1}{3}}{1 - ( x + \frac{1}{3} ) ( x - \frac{1}{3} )}

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

x + \frac{1}{3} + x - \frac{1}{3}

Group like terms

= x + x + \frac{1}{3} - \frac{1}{3}

Add similar elements:

x + x = 2 x

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

Add similar elements:

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

= 2 x

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

Expand

1 - (x + \frac{1}{3}) (x - \frac{1}{3}) : - x^{2} + \frac{10}{9}

1 - (x + \frac{1}{3}) (x - \frac{1}{3})

Expand

- (x + \frac{1}{3}) (x - \frac{1}{3}) : - x^{2} + \frac{1}{9}

Expand

(x + \frac{1}{3}) (x - \frac{1}{3}) : x^{2} - \frac{1}{9}

(x + \frac{1}{3}) (x - \frac{1}{3})

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = x, b = \frac{1}{3}

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

(\frac{1}{3})^{2} = \frac{1}{9}

(\frac{1}{3})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

3^{2} = 9

= \frac{1}{9}

= x^{2} - \frac{1}{9}

= - (x^{2} - \frac{1}{9})

Distribute parentheses

= - (x^{2}) - (- \frac{1}{9})

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - x^{2} + \frac{1}{9}

= 1 - x^{2} + \frac{1}{9}

Combine the fractions

1 + \frac{1}{9} : \frac{10}{9}

1 + \frac{1}{9}

Convert element to fraction:

1 = \frac{1 \cdot 9}{9}

= \frac{1 \cdot 9}{9} + \frac{1}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 9 + 1}{9}

1 \cdot 9 + 1 = 10

1 \cdot 9 + 1

Multiply the numbers:

1 \cdot 9 = 9

= 9 + 1

Add the numbers:

9 + 1 = 10

= 10

= \frac{10}{9}

= - x^{2} + \frac{10}{9}

= \frac{2 x}{- x ^{2} + \frac{10}{9}}

Join

- x^{2} + \frac{10}{9} : \frac{- 9 x ^{2} + 10}{9}

- x^{2} + \frac{10}{9}

Convert element to fraction:

x^{2} = \frac{x ^{2} 9}{9}

= - \frac{x ^{2} \cdot 9}{9} + \frac{10}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{- x ^{2} \cdot 9 + 10}{9}

= \frac{2 x}{\frac{- 9 x ^{2} + 10}{9}}

Apply the fraction rule:

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

= \frac{2 x \cdot 9}{- x ^{2} \cdot 9 + 10}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{18 x}{- 9 x ^{2} + 10}

\frac{18 x}{- 9 x ^{2} + 10} = 2

Multiply both sides by

- 9 x^{2} + 10

\frac{18 x}{- 9 x ^{2} + 10} = 2

Multiply both sides by

- 9 x^{2} + 10

\frac{18 x}{- 9 x ^{2} + 10} (- 9 x^{2} + 10) = 2 (- 9 x^{2} + 10)

Simplify

18 x = 2 (- 9 x^{2} + 10)

18 x = 2 (- 9 x^{2} + 10)

Solve

18 x = 2 (- 9 x^{2} + 10) : x = - \frac{5}{3}, x = \frac{2}{3}

18 x = 2 (- 9 x^{2} + 10)

Expand

2 (- 9 x^{2} + 10) : - 18 x^{2} + 20

2 (- 9 x^{2} + 10)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = - 9 x^{2}, c = 10

= 2 (- 9 x^{2}) + 2 \cdot 10

Apply minus-plus rules

+ (- a) = - a

= - 2 \cdot 9 x^{2} + 2 \cdot 10

Simplify

- 2 \cdot 9 x^{2} + 2 \cdot 10 : - 18 x^{2} + 20

- 2 \cdot 9 x^{2} + 2 \cdot 10

Multiply the numbers:

2 \cdot 9 = 18

= - 18 x^{2} + 2 \cdot 10

Multiply the numbers:

2 \cdot 10 = 20

= - 18 x^{2} + 20

= - 18 x^{2} + 20

18 x = - 18 x^{2} + 20

Switch sides

- 18 x^{2} + 20 = 18 x

Move

18 x

to the left side

- 18 x^{2} + 20 = 18 x

Subtract

18 x

from both sides

- 18 x^{2} + 20 - 18 x = 18 x - 18 x

Simplify

- 18 x^{2} + 20 - 18 x = 0

- 18 x^{2} + 20 - 18 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- 18 x^{2} - 18 x + 20 = 0

Solve with the quadratic formula

- 18 x^{2} - 18 x + 20 = 0

Quadratic Equation Formula:

For

a = - 18, b = - 18, c = 20

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 ( - 18 ) \cdot 20}{2 ( - 18 )}

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 ( - 18 ) \cdot 20}{2 ( - 18 )}

(- 18)^{2} - 4 (- 18) \cdot 20 = 42

(- 18)^{2} - 4 (- 18) \cdot 20

Apply rule

- (- a) = a

= (- 18)^{2} + 4 \cdot 18 \cdot 20

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 18)^{2} = 1 8^{2}

= 1 8^{2} + 4 \cdot 18 \cdot 20

Multiply the numbers:

4 \cdot 18 \cdot 20 = 1440

= 1 8^{2} + 1440

1 8^{2} = 324

= 324 + 1440

Add the numbers:

324 + 1440 = 1764

= 1764

Factor the number:

1764 = 4 2^{2}

= 4 2^{2}

Apply radical rule:

4 2^{2} = 42

= 42

x_{1, 2} = \frac{- ( - 18 ) \pm 42}{2 ( - 18 )}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 42}{2 ( - 18 )}, x_{2} = \frac{- ( - 18 ) - 42}{2 ( - 18 )}

x = \frac{- ( - 18 ) + 42}{2 ( - 18 )} : - \frac{5}{3}

\frac{- ( - 18 ) + 42}{2 ( - 18 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{18 + 42}{- 2 \cdot 18}

Add the numbers:

18 + 42 = 60

= \frac{60}{- 2 \cdot 18}

Multiply the numbers:

2 \cdot 18 = 36

= \frac{60}{- 36}

Apply the fraction rule:

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

= - \frac{60}{36}

Cancel the common factor:

12

= - \frac{5}{3}

x = \frac{- ( - 18 ) - 42}{2 ( - 18 )} : \frac{2}{3}

\frac{- ( - 18 ) - 42}{2 ( - 18 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{18 - 42}{- 2 \cdot 18}

Subtract the numbers:

18 - 42 = - 24

= \frac{- 24}{- 2 \cdot 18}

Multiply the numbers:

2 \cdot 18 = 36

= \frac{- 24}{- 36}

Apply the fraction rule:

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

= \frac{24}{36}

Cancel the common factor:

12

= \frac{2}{3}

The solutions to the quadratic equation are:

x = - \frac{5}{3}, x = \frac{2}{3}

x = - \frac{5}{3}, x = \frac{2}{3}

Verify Solutions

Find undefined (singularity) points:

x = - \frac{10}{3}, x = \frac{10}{3}

Take the denominator(s) of

\frac{x + \frac{1}{3} + x - \frac{1}{3}}{1 - ( x + \frac{1}{3} ) ( x - \frac{1}{3} )}

and compare to zero

Solve

1 - (x + \frac{1}{3}) (x - \frac{1}{3}) = 0 : x = - \frac{10}{3}, x = \frac{10}{3}

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

Expand

1 - (x + \frac{1}{3}) (x - \frac{1}{3}) : - x^{2} + \frac{10}{9}

1 - (x + \frac{1}{3}) (x - \frac{1}{3})

Expand

- (x + \frac{1}{3}) (x - \frac{1}{3}) : - x^{2} + \frac{1}{9}

Expand

(x + \frac{1}{3}) (x - \frac{1}{3}) : x^{2} - \frac{1}{9}

(x + \frac{1}{3}) (x - \frac{1}{3})

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = x, b = \frac{1}{3}

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

(\frac{1}{3})^{2} = \frac{1}{9}

(\frac{1}{3})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

3^{2} = 9

= \frac{1}{9}

= x^{2} - \frac{1}{9}

= - (x^{2} - \frac{1}{9})

Distribute parentheses

= - (x^{2}) - (- \frac{1}{9})

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - x^{2} + \frac{1}{9}

= 1 - x^{2} + \frac{1}{9}

Combine the fractions

1 + \frac{1}{9} : \frac{10}{9}

1 + \frac{1}{9}

Convert element to fraction:

1 = \frac{1 \cdot 9}{9}

= \frac{1 \cdot 9}{9} + \frac{1}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 9 + 1}{9}

1 \cdot 9 + 1 = 10

1 \cdot 9 + 1

Multiply the numbers:

1 \cdot 9 = 9

= 9 + 1

Add the numbers:

9 + 1 = 10

= 10

= \frac{10}{9}

= - x^{2} + \frac{10}{9}

- x^{2} + \frac{10}{9} = 0

Solve with the quadratic formula

- x^{2} + \frac{10}{9} = 0

Quadratic Equation Formula:

For

a = - 1, b = 0, c = \frac{10}{9}

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \frac{10}{9}}{2 ( - 1 )}

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \frac{10}{9}}{2 ( - 1 )}

0^{2} - 4 (- 1) \frac{10}{9} = \frac{2 10}{3}

0^{2} - 4 (- 1) \frac{10}{9}

Apply rule

0^{a} = 0

0^{2} = 0

= 0 - 4 (- 1) \frac{10}{9}

Apply rule

- (- a) = a

= 0 + 4 \cdot 1 \cdot \frac{10}{9}

4 \cdot 1 \cdot \frac{10}{9} = \frac{40}{9}

4 \cdot 1 \cdot \frac{10}{9}

Multiply fractions:

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

= 1 \cdot \frac{10 \cdot 4}{9}

Multiply the numbers:

10 \cdot 4 = 40

= 1 \cdot \frac{40}{9}

Multiply:

1 \cdot \frac{40}{9} = \frac{40}{9}

= \frac{40}{9}

= 0 + \frac{40}{9}

0 + \frac{40}{9} = \frac{40}{9}

= \frac{40}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{40}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{40}{3}

40 = 210

40

Prime factorization of

40 : 2^{3} \cdot 5

40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

Apply exponent rule:

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

= 2^{2} \cdot 2 \cdot 5

Apply radical rule:

= 2^{2} 2 \cdot 5

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 5

Refine

= 210

= \frac{2 10}{3}

x_{1, 2} = \frac{- 0 \pm \frac{2 10}{3}}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 0 + \frac{2 10}{3}}{2 ( - 1 )}, x_{2} = \frac{- 0 - \frac{2 10}{3}}{2 ( - 1 )}

x = \frac{- 0 + \frac{2 10}{3}}{2 ( - 1 )} : - \frac{10}{3}

\frac{- 0 + \frac{2 10}{3}}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 0 + \frac{2 10}{3}}{- 2 \cdot 1}

- 0 + \frac{2 10}{3} = \frac{2 10}{3}

= \frac{\frac{2 10}{3}}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{\frac{2 10}{3}}{- 2}

Apply the fraction rule:

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

= - \frac{\frac{2 10}{3}}{2}

Apply the fraction rule:

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

\frac{\frac{2 10}{3}}{2} = \frac{2 10}{3 \cdot 2}

= - \frac{2 10}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= - \frac{2 10}{6}

Cancel the common factor:

2

= - \frac{10}{3}

x = \frac{- 0 - \frac{2 10}{3}}{2 ( - 1 )} : \frac{10}{3}

\frac{- 0 - \frac{2 10}{3}}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 0 - \frac{2 10}{3}}{- 2 \cdot 1}

- 0 - \frac{2 10}{3} = - \frac{2 10}{3}

= \frac{- \frac{2 10}{3}}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- \frac{2 10}{3}}{- 2}

Apply the fraction rule:

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

= \frac{\frac{2 10}{3}}{2}

Apply the fraction rule:

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

= \frac{2 10}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 10}{6}

Cancel the common factor:

2

= \frac{10}{3}

The solutions to the quadratic equation are:

x = - \frac{10}{3}, x = \frac{10}{3}

The following points are undefined

x = - \frac{10}{3}, x = \frac{10}{3}

Combine undefined points with solutions:

x = - \frac{5}{3}, x = \frac{2}{3}

x = - \frac{5}{3}, x = \frac{2}{3}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

Check the solution

- \frac{5}{3} :

False

- \frac{5}{3}

Plug in

n = 1

- \frac{5}{3}

For

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

plug in

x = - \frac{5}{3}

arctan (- \frac{5}{3} + \frac{1}{3}) + arctan (- \frac{5}{3} - \frac{1}{3}) = arctan (2)

Refine

- 2.03444 \dots = 1.10714 \dots

\Rightarrow False

Check the solution

\frac{2}{3} :

True

\frac{2}{3}

Plug in

n = 1

\frac{2}{3}

For

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

plug in

x = \frac{2}{3}

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

Refine

1.10714 \dots = 1.10714 \dots

\Rightarrow True

x = \frac{2}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for arctan(x+1/3)+arctan(x-1/3)=arctan(2) ?
The general solution for arctan(x+1/3)+arctan(x-1/3)=arctan(2) is x= 2/3