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

The general solution for arctan(2x)+arctan(3x)= pi/4 is x= 1/6

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arctan(2x)+arctan(3x)= pi/4

Solution

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

Solution

x = \frac{1}{6}

Solution steps

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

Rewrite using trig identities

arctan (2 x) + arctan (3 x)

Use the Sum to Product identity:

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

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

arctan (\frac{2 x + 3 x}{1 - 2 x \cdot 3 x}) = \frac{π}{4}

Apply trig inverse properties

arctan (\frac{2 x + 3 x}{1 - 2 x \cdot 3 x}) = \frac{π}{4}

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

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

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

Solve

\frac{2 x + 3 x}{1 - 2 x \cdot 3 x} = 1 : x = - 1, x = \frac{1}{6}

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

Simplify

\frac{2 x + 3 x}{1 - 2 x \cdot 3 x} : \frac{5 x}{1 - 6 x ^{2}}

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

Add similar elements:

2 x + 3 x = 5 x

= \frac{5 x}{1 - 2 \cdot 3 xx}

1 - 2 x \cdot 3 x = 1 - 6 x^{2}

1 - 2 x \cdot 3 x

2 x \cdot 3 x = 6 x^{2}

2 x \cdot 3 x

Multiply the numbers:

2 \cdot 3 = 6

= 6 xx

Apply exponent rule:

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

xx = x^{1 + 1}

= 6 x^{1 + 1}

Add the numbers:

1 + 1 = 2

= 6 x^{2}

= 1 - 6 x^{2}

= \frac{5 x}{1 - 6 x ^{2}}

\frac{5 x}{1 - 6 x ^{2}} = 1

Multiply both sides by

1 - 6 x^{2}

\frac{5 x}{1 - 6 x ^{2}} = 1

Multiply both sides by

1 - 6 x^{2}

\frac{5 x}{1 - 6 x ^{2}} (1 - 6 x^{2}) = 1 \cdot (1 - 6 x^{2})

Simplify

\frac{5 x}{1 - 6 x ^{2}} (1 - 6 x^{2}) = 1 \cdot (1 - 6 x^{2})

Simplify

\frac{5 x}{1 - 6 x ^{2}} (1 - 6 x^{2}) : 5 x

\frac{5 x}{1 - 6 x ^{2}} (1 - 6 x^{2})

Multiply fractions:

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

= \frac{5 x ( 1 - 6 x ^{2} )}{1 - 6 x ^{2}}

Cancel the common factor:

1 - 6 x^{2}

= 5 x

Simplify

1 \cdot (1 - 6 x^{2}) : 1 - 6 x^{2}

1 \cdot (1 - 6 x^{2})

Multiply:

1 \cdot (1 - 6 x^{2}) = (1 - 6 x^{2})

= (1 - 6 x^{2})

Remove parentheses:

(a) = a

= 1 - 6 x^{2}

5 x = 1 - 6 x^{2}

5 x = 1 - 6 x^{2}

5 x = 1 - 6 x^{2}

Solve

5 x = 1 - 6 x^{2} : x = - 1, x = \frac{1}{6}

5 x = 1 - 6 x^{2}

Switch sides

1 - 6 x^{2} = 5 x

Move

5 x

to the left side

1 - 6 x^{2} = 5 x

Subtract

5 x

from both sides

1 - 6 x^{2} - 5 x = 5 x - 5 x

Simplify

1 - 6 x^{2} - 5 x = 0

1 - 6 x^{2} - 5 x = 0

Write in the standard form

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

- 6 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

- 6 x^{2} - 5 x + 1 = 0

Quadratic Equation Formula:

For

a = - 6, b = - 5, c = 1

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 6 ) \cdot 1}{2 ( - 6 )}

(- 5)^{2} - 4 (- 6) \cdot 1 = 7

(- 5)^{2} - 4 (- 6) \cdot 1

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 6 \cdot 1

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

x_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 7}{2 ( - 6 )}, x_{2} = \frac{- ( - 5 ) - 7}{2 ( - 6 )}

x = \frac{- ( - 5 ) + 7}{2 ( - 6 )} : - 1

\frac{- ( - 5 ) + 7}{2 ( - 6 )}

Remove parentheses:

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

= \frac{5 + 7}{- 2 \cdot 6}

Add the numbers:

5 + 7 = 12

= \frac{12}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{12}{- 12}

Apply the fraction rule:

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

= - \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= - 1

x = \frac{- ( - 5 ) - 7}{2 ( - 6 )} : \frac{1}{6}

\frac{- ( - 5 ) - 7}{2 ( - 6 )}

Remove parentheses:

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

= \frac{5 - 7}{- 2 \cdot 6}

Subtract the numbers:

5 - 7 = - 2

= \frac{- 2}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 2}{- 12}

Apply the fraction rule:

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

= \frac{2}{12}

Cancel the common factor:

2

= \frac{1}{6}

The solutions to the quadratic equation are:

x = - 1, x = \frac{1}{6}

x = - 1, x = \frac{1}{6}

Verify Solutions

Find undefined (singularity) points:

x = \frac{1}{6}, x = - \frac{1}{6}

Take the denominator(s) of

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

and compare to zero

Solve

1 - 2 x \cdot 3 x = 0 : x = \frac{1}{6}, x = - \frac{1}{6}

1 - 2 x \cdot 3 x = 0

Move

1

to the right side

1 - 2 x \cdot 3 x = 0

Subtract

1

from both sides

1 - 2 x \cdot 3 x - 1 = 0 - 1

Simplify

- 2 x \cdot 3 x = - 1

- 2 x \cdot 3 x = - 1

Simplify

- 6 x^{2} = - 1

Divide both sides by

- 6

\frac{- 6 x ^{2}}{- 6} = \frac{- 1}{- 6}

x^{2} = \frac{1}{6}

For

x^{2} = f (a)

the solutions are

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

x = \frac{1}{6}, x = - \frac{1}{6}

\frac{1}{6} = \frac{1}{6}

\frac{1}{6}

Apply radical rule:

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

= \frac{1}{6}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{6}

- \frac{1}{6} = - \frac{1}{6}

- \frac{1}{6}

Apply radical rule:

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

= - \frac{1}{6}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{6}

x = \frac{1}{6}, x = - \frac{1}{6}

The following points are undefined

x = \frac{1}{6}, x = - \frac{1}{6}

Combine undefined points with solutions:

x = - 1, x = \frac{1}{6}

x = - 1, x = \frac{1}{6}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

- 1 :

False

- 1

Plug in

n = 1

- 1

For

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

plug in

x = - 1

arctan (2 (- 1)) + arctan (3 (- 1)) = \frac{π}{4}

Refine

- 2.35619 \dots = 0.78539 \dots

\Rightarrow False

Check the solution

\frac{1}{6} :

True

\frac{1}{6}

Plug in

n = 1

\frac{1}{6}

For

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

plug in

x = \frac{1}{6}

arctan (2 \cdot \frac{1}{6}) + arctan (3 \cdot \frac{1}{6}) = \frac{π}{4}

Refine

0.78539 \dots = 0.78539 \dots

\Rightarrow True

x = \frac{1}{6}

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Frequently Asked Questions (FAQ)

What is the general solution for arctan(2x)+arctan(3x)= pi/4 ?
The general solution for arctan(2x)+arctan(3x)= pi/4 is x= 1/6