What is the general solution for arctan(3x)+arctan(x)= pi/4 ?

The general solution for arctan(3x)+arctan(x)= pi/4 is x=(sqrt(7)-2)/3

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Popular Trigonometry >

arctan(3x)+arctan(x)= pi/4

Solution

arctan (3 x) + arctan (x) = \frac{π}{4}

Solution

x = \frac{7 - 2}{3}

Solution steps

arctan (3 x) + arctan (x) = \frac{π}{4}

Rewrite using trig identities

arctan (3 x) + arctan (x)

Use the Sum to Product identity:

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

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

arctan (\frac{3 x + x}{1 - 3 xx}) = \frac{π}{4}

Apply trig inverse properties

arctan (\frac{3 x + x}{1 - 3 xx}) = \frac{π}{4}

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

\frac{3 x + x}{1 - 3 xx} = tan (\frac{π}{4})

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

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}

= 1

= 1

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

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

Solve

\frac{3 x + x}{1 - 3 xx} = 1 : x = - \frac{2 + 7}{3}, x = \frac{7 - 2}{3}

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

Simplify

\frac{3 x + x}{1 - 3 xx} : \frac{4 x}{1 - 3 x ^{2}}

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

Add similar elements:

3 x + x = 4 x

= \frac{4 x}{1 - 3 xx}

1 - 3 xx = 1 - 3 x^{2}

1 - 3 xx

3 xx = 3 x^{2}

3 xx

Apply exponent rule:

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

xx = x^{1 + 1}

= 3 x^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3 x^{2}

= 1 - 3 x^{2}

= \frac{4 x}{1 - 3 x ^{2}}

\frac{4 x}{1 - 3 x ^{2}} = 1

Multiply both sides by

1 - 3 x^{2}

\frac{4 x}{1 - 3 x ^{2}} = 1

Multiply both sides by

1 - 3 x^{2}

\frac{4 x}{1 - 3 x ^{2}} (1 - 3 x^{2}) = 1 \cdot (1 - 3 x^{2})

Simplify

\frac{4 x}{1 - 3 x ^{2}} (1 - 3 x^{2}) = 1 \cdot (1 - 3 x^{2})

Simplify

\frac{4 x}{1 - 3 x ^{2}} (1 - 3 x^{2}) : 4 x

\frac{4 x}{1 - 3 x ^{2}} (1 - 3 x^{2})

Multiply fractions:

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

= \frac{4 x ( 1 - 3 x ^{2} )}{1 - 3 x ^{2}}

Cancel the common factor:

1 - 3 x^{2}

= 4 x

Simplify

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

1 \cdot (1 - 3 x^{2})

Multiply:

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

= (1 - 3 x^{2})

Remove parentheses:

(a) = a

= 1 - 3 x^{2}

4 x = 1 - 3 x^{2}

4 x = 1 - 3 x^{2}

4 x = 1 - 3 x^{2}

Solve

4 x = 1 - 3 x^{2} : x = - \frac{2 + 7}{3}, x = \frac{7 - 2}{3}

4 x = 1 - 3 x^{2}

Switch sides

1 - 3 x^{2} = 4 x

Move

4 x

to the left side

1 - 3 x^{2} = 4 x

Subtract

4 x

from both sides

1 - 3 x^{2} - 4 x = 4 x - 4 x

Simplify

1 - 3 x^{2} - 4 x = 0

1 - 3 x^{2} - 4 x = 0

Write in the standard form

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

- 3 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

- 3 x^{2} - 4 x + 1 = 0

Quadratic Equation Formula:

For

a = - 3, b = - 4, c = 1

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 3 ) \cdot 1}{2 ( - 3 )}

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 3 ) \cdot 1}{2 ( - 3 )}

(- 4)^{2} - 4 (- 3) \cdot 1 = 27

(- 4)^{2} - 4 (- 3) \cdot 1

Apply rule

- (- a) = a

= (- 4)^{2} + 4 \cdot 3 \cdot 1

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} + 4 \cdot 3 \cdot 1

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 4^{2} + 12

4^{2} = 16

= 16 + 12

Add the numbers:

16 + 12 = 28

= 28

Prime factorization of

28 : 2^{2} \cdot 7

28

28

divides by

228 = 14 \cdot 2

= 2 \cdot 14

14

divides by

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

Apply radical rule:

= 7 2^{2}

Apply radical rule:

2^{2} = 2

= 27

x_{1, 2} = \frac{- ( - 4 ) \pm 2 7}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 7}{2 ( - 3 )}, x_{2} = \frac{- ( - 4 ) - 2 7}{2 ( - 3 )}

x = \frac{- ( - 4 ) + 2 7}{2 ( - 3 )} : - \frac{2 + 7}{3}

\frac{- ( - 4 ) + 2 7}{2 ( - 3 )}

Remove parentheses:

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

= \frac{4 + 2 7}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4 + 2 7}{- 6}

Apply the fraction rule:

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

= - \frac{4 + 2 7}{6}

Cancel

\frac{4 + 2 7}{6} : \frac{2 + 7}{3}

\frac{4 + 2 7}{6}

Factor

4 + 27 : 2 (2 + 7)

4 + 27

Rewrite as

= 2 \cdot 2 + 27

Factor out common term

2

= 2 (2 + 7)

= \frac{2 ( 2 + 7 )}{6}

Cancel the common factor:

2

= \frac{2 + 7}{3}

= - \frac{2 + 7}{3}

x = \frac{- ( - 4 ) - 2 7}{2 ( - 3 )} : \frac{7 - 2}{3}

\frac{- ( - 4 ) - 2 7}{2 ( - 3 )}

Remove parentheses:

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

= \frac{4 - 2 7}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4 - 2 7}{- 6}

Apply the fraction rule:

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

4 - 27 = - (27 - 4)

= \frac{2 7 - 4}{6}

Factor

27 - 4 : 2 (7 - 2)

27 - 4

Rewrite as

= 27 - 2 \cdot 2

Factor out common term

2

= 2 (7 - 2)

= \frac{2 ( 7 - 2 )}{6}

Cancel the common factor:

2

= \frac{7 - 2}{3}

The solutions to the quadratic equation are:

x = - \frac{2 + 7}{3}, x = \frac{7 - 2}{3}

x = - \frac{2 + 7}{3}, x = \frac{7 - 2}{3}

Verify Solutions

Find undefined (singularity) points:

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

Take the denominator(s) of

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

and compare to zero

Solve

1 - 3 xx = 0 : x = \frac{1}{3}, x = - \frac{1}{3}

1 - 3 xx = 0

Move

1

to the right side

1 - 3 xx = 0

Subtract

1

from both sides

1 - 3 xx - 1 = 0 - 1

Simplify

- 3 xx = - 1

- 3 xx = - 1

Simplify

- 3 x^{2} = - 1

Divide both sides by

- 3

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

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

\frac{1}{3}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{3}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{3}

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

- \frac{1}{3}

Apply radical rule:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{3}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{3}

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

The following points are undefined

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

Combine undefined points with solutions:

x = - \frac{2 + 7}{3}, x = \frac{7 - 2}{3}

x = - \frac{2 + 7}{3}, x = \frac{7 - 2}{3}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arctan (3 x) + arctan (x) = \frac{π}{4}

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

Check the solution

- \frac{2 + 7}{3} :

False

- \frac{2 + 7}{3}

Plug in

n = 1

- \frac{2 + 7}{3}

For

arctan (3 x) + arctan (x) = \frac{π}{4}

plug in

x = - \frac{2 + 7}{3}

arctan (3 (- \frac{2 + 7}{3})) + arctan (- \frac{2 + 7}{3}) = \frac{π}{4}

Refine

- 2.35619 \dots = 0.78539 \dots

\Rightarrow False

Check the solution

\frac{7 - 2}{3} :

True

\frac{7 - 2}{3}

Plug in

n = 1

\frac{7 - 2}{3}

For

arctan (3 x) + arctan (x) = \frac{π}{4}

plug in

x = \frac{7 - 2}{3}

arctan (3 \cdot \frac{7 - 2}{3}) + arctan (\frac{7 - 2}{3}) = \frac{π}{4}

Refine

0.78539 \dots = 0.78539 \dots

\Rightarrow True

x = \frac{7 - 2}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for arctan(3x)+arctan(x)= pi/4 ?
The general solution for arctan(3x)+arctan(x)= pi/4 is x=(sqrt(7)-2)/3