What is the general solution for arctan(4-2x)=arctan(2x) ?

The general solution for arctan(4-2x)=arctan(2x) is x=1

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arctan(4-2x)=arctan(2x)

Solution

arctan (4 - 2 x) = arctan (2 x)

Solution

x = 1

Solution steps

arctan (4 - 2 x) = arctan (2 x)

Subtract

arctan (2 x)

from both sides

arctan (4 - 2 x) - arctan (2 x) = 0

Rewrite using trig identities

arctan (4 - 2 x) - arctan (2 x)

Use the Sum to Product identity:

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

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

arctan (\frac{4 - 2 x - 2 x}{1 + ( 4 - 2 x ) \cdot 2 x}) = 0

Apply trig inverse properties

arctan (\frac{4 - 2 x - 2 x}{1 + ( 4 - 2 x ) \cdot 2 x}) = 0

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

\frac{4 - 2 x - 2 x}{1 + ( 4 - 2 x ) \cdot 2 x} = tan (0)

tan (0) = 0

tan (0)

Use the following trivial identity:

tan (0) = 0

tan (0)

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}

= 0

= 0

\frac{4 - 2 x - 2 x}{1 + ( 4 - 2 x ) \cdot 2 x} = 0

\frac{4 - 2 x - 2 x}{1 + ( 4 - 2 x ) \cdot 2 x} = 0

Solve

\frac{4 - 2 x - 2 x}{1 + ( 4 - 2 x ) \cdot 2 x} = 0 : x = 1

\frac{4 - 2 x - 2 x}{1 + ( 4 - 2 x ) \cdot 2 x} = 0

Simplify

\frac{4 - 2 x - 2 x}{1 + ( 4 - 2 x ) \cdot 2 x} : \frac{4 - 4 x}{1 + 2 x ( 4 - 2 x )}

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

Add similar elements:

- 2 x - 2 x = - 4 x

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

\frac{4 - 4 x}{1 + 2 x ( 4 - 2 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 - 4 x = 0

Move

4

to the right side

4 - 4 x = 0

Subtract

4

from both sides

4 - 4 x - 4 = 0 - 4

Simplify

- 4 x = - 4

- 4 x = - 4

Divide both sides by

- 4

- 4 x = - 4

Divide both sides by

- 4

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

Simplify

x = 1

x = 1

Verify Solutions

Find undefined (singularity) points:

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

Take the denominator(s) of

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

and compare to zero

Solve

1 + (4 - 2 x) \cdot 2 x = 0 : x = - \frac{- 2 + 5}{2}, x = \frac{2 + 5}{2}

1 + (4 - 2 x) \cdot 2 x = 0

Expand

1 + (4 - 2 x) \cdot 2 x : 1 + 8 x - 4 x^{2}

1 + (4 - 2 x) \cdot 2 x

= 1 + 2 x (4 - 2 x)

Expand

2 x (4 - 2 x) : 8 x - 4 x^{2}

2 x (4 - 2 x)

Apply the distributive law:

a (b - c) = ab - a c

a = 2 x, b = 4, c = 2 x

= 2 x \cdot 4 - 2 x \cdot 2 x

= 2 \cdot 4 x - 2 \cdot 2 xx

Simplify

2 \cdot 4 x - 2 \cdot 2 xx : 8 x - 4 x^{2}

2 \cdot 4 x - 2 \cdot 2 xx

2 \cdot 4 x = 8 x

2 \cdot 4 x

Multiply the numbers:

2 \cdot 4 = 8

= 8 x

2 \cdot 2 xx = 4 x^{2}

2 \cdot 2 xx

Multiply the numbers:

2 \cdot 2 = 4

= 4 xx

Apply exponent rule:

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

xx = x^{1 + 1}

= 4 x^{1 + 1}

Add the numbers:

1 + 1 = 2

= 4 x^{2}

= 8 x - 4 x^{2}

= 8 x - 4 x^{2}

= 1 + 8 x - 4 x^{2}

1 + 8 x - 4 x^{2} = 0

Write in the standard form

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

- 4 x^{2} + 8 x + 1 = 0

Solve with the quadratic formula

- 4 x^{2} + 8 x + 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = 8, c = 1

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

8^{2} - 4 (- 4) \cdot 1 = 45

8^{2} - 4 (- 4) \cdot 1

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 8^{2} + 16

8^{2} = 64

= 64 + 16

Add the numbers:

64 + 16 = 80

= 80

Prime factorization of

80 : 2^{4} \cdot 5

80

80

divides by

280 = 40 \cdot 2

= 2 \cdot 40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 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 2 \cdot 5

= 2^{4} \cdot 5

= 2^{4} \cdot 5

Apply radical rule:

= 5 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 5

Refine

= 45

x_{1, 2} = \frac{- 8 \pm 4 5}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 8 + 4 5}{2 ( - 4 )}, x_{2} = \frac{- 8 - 4 5}{2 ( - 4 )}

x = \frac{- 8 + 4 5}{2 ( - 4 )} : - \frac{- 2 + 5}{2}

\frac{- 8 + 4 5}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 8 + 4 5}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8 + 4 5}{- 8}

Apply the fraction rule:

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

= - \frac{- 8 + 4 5}{8}

Cancel

\frac{- 8 + 4 5}{8} : \frac{5 - 2}{2}

\frac{- 8 + 4 5}{8}

Factor

- 8 + 45 : 4 (- 2 + 5)

- 8 + 45

Rewrite as

= - 4 \cdot 2 + 45

Factor out common term

4

= 4 (- 2 + 5)

= \frac{4 ( - 2 + 5 )}{8}

Cancel the common factor:

4

= \frac{- 2 + 5}{2}

= - \frac{5 - 2}{2}

= - \frac{- 2 + 5}{2}

x = \frac{- 8 - 4 5}{2 ( - 4 )} : \frac{2 + 5}{2}

\frac{- 8 - 4 5}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 8 - 4 5}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8 - 4 5}{- 8}

Apply the fraction rule:

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

- 8 - 45 = - (8 + 45)

= \frac{8 + 4 5}{8}

Factor

8 + 45 : 4 (2 + 5)

8 + 45

Rewrite as

= 4 \cdot 2 + 45

Factor out common term

4

= 4 (2 + 5)

= \frac{4 ( 2 + 5 )}{8}

Cancel the common factor:

4

= \frac{2 + 5}{2}

The solutions to the quadratic equation are:

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

The following points are undefined

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

Combine undefined points with solutions:

x = 1

x = 1

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arctan (4 - 2 x) = arctan (2 x)

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

Check the solution

1 :

True

1

Plug in

n = 1

1

For

arctan (4 - 2 x) = arctan (2 x)

plug in

x = 1

arctan (4 - 2 \cdot 1) = arctan (2 \cdot 1)

Refine

1.10714 \dots = 1.10714 \dots

\Rightarrow True

x = 1

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Frequently Asked Questions (FAQ)

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