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

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

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Popular Trigonometry >

arctan(x+2)-arctan(x+1)= pi/4

Solution

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

Solution

x = - 1, x = - 2

Solution steps

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

Rewrite using trig identities

arctan (x + 2) - arctan (x + 1)

Use the Sum to Product identity:

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

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

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

Apply trig inverse properties

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

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

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

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

Solve

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

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

Simplify

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

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

Expand

1 + (x + 2) (x + 1) : x^{2} + 3 x + 3

1 + (x + 2) (x + 1)

Expand

(x + 2) (x + 1) : x^{2} + 3 x + 2

(x + 2) (x + 1)

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = x, b = 2, c = x, d = 1

= xx + x \cdot 1 + 2 x + 2 \cdot 1

= xx + 1 \cdot x + 2 x + 2 \cdot 1

Simplify

xx + 1 \cdot x + 2 x + 2 \cdot 1 : x^{2} + 3 x + 2

xx + 1 \cdot x + 2 x + 2 \cdot 1

Add similar elements:

1 \cdot x + 2 x = 3 x

= xx + 3 x + 2 \cdot 1

xx = x^{2}

xx

Apply exponent rule:

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

xx = x^{1 + 1}

= x^{1 + 1}

Add the numbers:

1 + 1 = 2

= x^{2}

2 \cdot 1 = 2

2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

= x^{2} + 3 x + 2

= x^{2} + 3 x + 2

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

Simplify

1 + x^{2} + 3 x + 2 : x^{2} + 3 x + 3

1 + x^{2} + 3 x + 2

Group like terms

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

Add the numbers:

1 + 2 = 3

= x^{2} + 3 x + 3

= x^{2} + 3 x + 3

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

Expand

x + 2 - (x + 1) : 1

x + 2 - (x + 1)

- (x + 1) : - x - 1

- (x + 1)

Distribute parentheses

= - (x) - (1)

Apply minus-plus rules

+ (- a) = - a

= - x - 1

= x + 2 - x - 1

Simplify

x + 2 - x - 1 : 1

x + 2 - x - 1

Group like terms

= x - x + 2 - 1

Add similar elements:

x - x = 0

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= 1

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

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

Multiply both sides by

x^{2} + 3 x + 3

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

Multiply both sides by

x^{2} + 3 x + 3

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

x^{2} + 3 x + 3

= 1

Simplify

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

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

Multiply:

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

= (x^{2} + 3 x + 3)

Remove parentheses:

(a) = a

= x^{2} + 3 x + 3

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

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

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

Solve

1 = x^{2} + 3 x + 3 : x = - 1, x = - 2

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

Switch sides

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

Move

1

to the left side

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

Subtract

1

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 3, c = 2

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

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

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

x_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 1}{2 \cdot 1}, x_{2} = \frac{- 3 - 1}{2 \cdot 1}

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

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

x = - 1, x = - 2

x = - 1, x = - 2

x = - 1, x = - 2

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

- 1 :

True

- 1

Plug in

n = 1

- 1

For

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

plug in

x = - 1

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

Refine

0.78539 \dots = 0.78539 \dots

\Rightarrow True

Check the solution

- 2 :

True

- 2

Plug in

n = 1

- 2

For

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

plug in

x = - 2

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

Refine

0.78539 \dots = 0.78539 \dots

\Rightarrow True

x = - 1, x = - 2

Graph

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Frequently Asked Questions (FAQ)

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