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

The general solution for arctan(x)+arctan(2x)= pi/4 is x=(sqrt(17)-3)/4

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

Solution

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

Solution

x = \frac{17 - 3}{4}

Solution steps

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

Rewrite using trig identities

arctan (x) + arctan (2 x)

Use the Sum to Product identity:

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

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

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

Apply trig inverse properties

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

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

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

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

Solve

\frac{x + 2 x}{1 - x \cdot 2 x} = 1 : x = - \frac{3 + 17}{4}, x = \frac{17 - 3}{4}

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

Simplify

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

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

Add similar elements:

x + 2 x = 3 x

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

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

1 - x \cdot 2 x

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

x \cdot 2 x

Apply exponent rule:

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

xx = x^{1 + 1}

= 2 x^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2 x^{2}

= 1 - 2 x^{2}

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

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

Multiply both sides by

1 - 2 x^{2}

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

Multiply both sides by

1 - 2 x^{2}

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

1 - 2 x^{2}

= 3 x

Simplify

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

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

Multiply:

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

= (1 - 2 x^{2})

Remove parentheses:

(a) = a

= 1 - 2 x^{2}

3 x = 1 - 2 x^{2}

3 x = 1 - 2 x^{2}

3 x = 1 - 2 x^{2}

Solve

3 x = 1 - 2 x^{2} : x = - \frac{3 + 17}{4}, x = \frac{17 - 3}{4}

3 x = 1 - 2 x^{2}

Switch sides

1 - 2 x^{2} = 3 x

Move

3 x

to the left side

1 - 2 x^{2} = 3 x

Subtract

3 x

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 3, c = 1

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

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

(- 3)^{2} - 4 (- 2) \cdot 1 = 17

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} + 8

3^{2} = 9

= 9 + 8

Add the numbers:

9 + 8 = 17

= 17

x_{1, 2} = \frac{- ( - 3 ) \pm 17}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 17}{2 ( - 2 )}, x_{2} = \frac{- ( - 3 ) - 17}{2 ( - 2 )}

x = \frac{- ( - 3 ) + 17}{2 ( - 2 )} : - \frac{3 + 17}{4}

\frac{- ( - 3 ) + 17}{2 ( - 2 )}

Remove parentheses:

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

= \frac{3 + 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 + 17}{- 4}

Apply the fraction rule:

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

= - \frac{3 + 17}{4}

x = \frac{- ( - 3 ) - 17}{2 ( - 2 )} : \frac{17 - 3}{4}

\frac{- ( - 3 ) - 17}{2 ( - 2 )}

Remove parentheses:

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

= \frac{3 - 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 - 17}{- 4}

Apply the fraction rule:

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

3 - 17 = - (17 - 3)

= \frac{17 - 3}{4}

The solutions to the quadratic equation are:

x = - \frac{3 + 17}{4}, x = \frac{17 - 3}{4}

x = - \frac{3 + 17}{4}, x = \frac{17 - 3}{4}

Verify Solutions

Find undefined (singularity) points:

x = \frac{1}{2}, x = - \frac{1}{2}

Take the denominator(s) of

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

and compare to zero

Solve

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

1 - x \cdot 2 x = 0

Move

1

to the right side

1 - x \cdot 2 x = 0

Subtract

1

from both sides

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

Simplify

- x \cdot 2 x = - 1

- x \cdot 2 x = - 1

Simplify

- 2 x^{2} = - 1

Divide both sides by

- 2

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

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

For

x^{2} = f (a)

the solutions are

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

x = \frac{1}{2}, x = - \frac{1}{2}

\frac{1}{2} = \frac{1}{2}

\frac{1}{2}

Apply radical rule:

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

= \frac{1}{2}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{2}

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

- \frac{1}{2}

Apply radical rule:

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

= - \frac{1}{2}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{2}

x = \frac{1}{2}, x = - \frac{1}{2}

The following points are undefined

x = \frac{1}{2}, x = - \frac{1}{2}

Combine undefined points with solutions:

x = - \frac{3 + 17}{4}, x = \frac{17 - 3}{4}

x = - \frac{3 + 17}{4}, x = \frac{17 - 3}{4}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

- \frac{3 + 17}{4} :

False

- \frac{3 + 17}{4}

Plug in

n = 1

- \frac{3 + 17}{4}

For

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

plug in

x = - \frac{3 + 17}{4}

arctan (- \frac{3 + 17}{4}) + arctan (2 (- \frac{3 + 17}{4})) = \frac{π}{4}

Refine

- 2.35619 \dots = 0.78539 \dots

\Rightarrow False

Check the solution

\frac{17 - 3}{4} :

True

\frac{17 - 3}{4}

Plug in

n = 1

\frac{17 - 3}{4}

For

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

plug in

x = \frac{17 - 3}{4}

arctan (\frac{17 - 3}{4}) + arctan (2 \cdot \frac{17 - 3}{4}) = \frac{π}{4}

Refine

0.78539 \dots = 0.78539 \dots

\Rightarrow True

x = \frac{17 - 3}{4}

Popular Examples

1+cos(θ)=(2+sqrt(3))/2

1 + cos (θ) = \frac{2 + 3}{2}

1+3sin(x)=2

1 + 3 sin (x) = 2

cos(x/2+pi/3)= 1/(sqrt(2))

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

sinh(x)=(sqrt(2))/2

sinh (x) = \frac{2}{2}

sin^2(2x)=cos^2(x)

sin^{2} (2 x) = cos^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for arctan(x)+arctan(2x)= pi/4 ?
The general solution for arctan(x)+arctan(2x)= pi/4 is x=(sqrt(17)-3)/4