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

The general solution for arccos(x)+arccos(2x)= pi/2 is x= 1/(sqrt(5))

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Popular Trigonometry >

arccos(x)+arccos(2x)= pi/2

Solution

arccos (x) + arccos (2 x) = \frac{π}{2}

Solution

x = \frac{1}{5}

Solution steps

arccos (x) + arccos (2 x) = \frac{π}{2}

Rewrite using trig identities

arccos (x) + arccos (2 x)

Use the Sum to Product identity:

arccos (s) + arccos (t) = arccos (s t - (1 - s^{2}) (1 - t^{2}))

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

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

Apply trig inverse properties

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

arccos (x) = a \Rightarrow x = cos (a)

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

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= 0

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

Solve

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

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

Expand

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) : 2 x^{2} - 1 - 5 x^{2} + 4 x^{4}

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

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}

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

Expand

2 x^{2} - (1 - x^{2}) (1 - (2 x)^{2}) : 2 x^{2} - 1 - 5 x^{2} + 4 x^{4}

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

(1 - x^{2}) (1 - (2 x)^{2}) = 1 - 5 x^{2} + 4 x^{4}

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

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

(2 x)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} x^{2}

2^{2} = 4

= 4 x^{2}

= (- x^{2} + 1) (- 4 x^{2} + 1)

Expand

(1 - x^{2}) (1 - 4 x^{2}) : 1 - 5 x^{2} + 4 x^{4}

(1 - x^{2}) (1 - 4 x^{2})

Apply FOIL method:

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

a = 1, b = - x^{2}, c = 1, d = - 4 x^{2}

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

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= 1 \cdot 1 - 1 \cdot 4 x^{2} - 1 \cdot x^{2} + 4 x^{2} x^{2}

Simplify

1 \cdot 1 - 1 \cdot 4 x^{2} - 1 \cdot x^{2} + 4 x^{2} x^{2} : 1 - 5 x^{2} + 4 x^{4}

1 \cdot 1 - 1 \cdot 4 x^{2} - 1 \cdot x^{2} + 4 x^{2} x^{2}

1 \cdot 1 = 1

1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 1

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

1 \cdot 4 x^{2}

Multiply the numbers:

1 \cdot 4 = 4

= 4 x^{2}

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

1 \cdot x^{2}

Multiply:

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

= x^{2}

4 x^{2} x^{2} = 4 x^{4}

4 x^{2} x^{2}

Apply exponent rule:

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

x^{2} x^{2} = x^{2 + 2}

= 4 x^{2 + 2}

Add the numbers:

2 + 2 = 4

= 4 x^{4}

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

Add similar elements:

- 4 x^{2} - x^{2} = - 5 x^{2}

= 1 - 5 x^{2} + 4 x^{4}

= 1 - 5 x^{2} + 4 x^{4}

= 1 - 5 x^{2} + 4 x^{4}

= 2 x^{2} - 4 x^{4} - 5 x^{2} + 1

= 2 x^{2} - 1 - 5 x^{2} + 4 x^{4}

2 x^{2} - 1 - 5 x^{2} + 4 x^{4} = 0

Remove square roots

2 x^{2} - 1 - 5 x^{2} + 4 x^{4} = 0

Subtract

2 x^{2}

from both sides

2 x^{2} - 1 - 5 x^{2} + 4 x^{4} - 2 x^{2} = 0 - 2 x^{2}

Simplify

- 1 - 5 x^{2} + 4 x^{4} = - 2 x^{2}

Square both sides:

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

2 x^{2} - 1 - 5 x^{2} + 4 x^{4} = 0

(- 1 - 5 x^{2} + 4 x^{4})^{2} = (- 2 x^{2})^{2}

Expand

(- 1 - 5 x^{2} + 4 x^{4})^{2} : 1 - 5 x^{2} + 4 x^{4}

(- 1 - 5 x^{2} + 4 x^{4})^{2}

Apply exponent rule:

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

n

is even

(- 1 - 5 x^{2} + 4 x^{4})^{2} = (1 - 5 x^{2} + 4 x^{4})^{2}

= (1 - 5 x^{2} + 4 x^{4})^{2}

Apply radical rule:

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

= ((1 - 5 x^{2} + 4 x^{4})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 - 5 x^{2} + 4 x^{4}

Expand

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

(- 2 x^{2})^{2}

Apply exponent rule:

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

n

is even

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= x^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= x^{4}

= 2^{2} x^{4}

2^{2} = 4

= 4 x^{4}

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

Solve

1 - 5 x^{2} + 4 x^{4} = 4 x^{4} : x = \frac{1}{5}, x = - \frac{1}{5}

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

Move

1

to the right side

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

Subtract

1

from both sides

1 - 5 x^{2} + 4 x^{4} - 1 = 4 x^{4} - 1

Simplify

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

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

Move

4 x^{4}

to the left side

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

Subtract

4 x^{4}

from both sides

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

Simplify

- 5 x^{2} = - 1

- 5 x^{2} = - 1

Divide both sides by

- 5

- 5 x^{2} = - 1

Divide both sides by

- 5

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

x = \frac{1}{5}, x = - \frac{1}{5}

\frac{1}{5} = \frac{1}{5}

\frac{1}{5}

Apply radical rule:

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

= \frac{1}{5}

Apply radical rule:

1 = 1

1 = 1

= \frac{1}{5}

- \frac{1}{5} = - \frac{1}{5}

- \frac{1}{5}

Apply radical rule:

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

= - \frac{1}{5}

Apply radical rule:

1 = 1

1 = 1

= - \frac{1}{5}

x = \frac{1}{5}, x = - \frac{1}{5}

x = \frac{1}{5}, x = - \frac{1}{5}

Verify Solutions:

x = \frac{1}{5}

True

, x = - \frac{1}{5}

True

Check the solutions by plugging them into

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

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

Plug in

x = \frac{1}{5} :

True

(\frac{1}{5}) \cdot 2 (\frac{1}{5}) - (1 - (\frac{1}{5})^{2}) (1 - (2 (\frac{1}{5}))^{2}) = 0

(\frac{1}{5}) \cdot 2 (\frac{1}{5}) - (1 - (\frac{1}{5})^{2}) (1 - (2 (\frac{1}{5}))^{2}) = 0

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

Remove parentheses:

(a) = a

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

\frac{1}{5} \cdot 2 \cdot \frac{1}{5} = \frac{2}{5}

\frac{1}{5} \cdot 2 \cdot \frac{1}{5}

Multiply fractions:

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

= \frac{1 \cdot 1 \cdot 2}{5 5}

Multiply the numbers:

1 \cdot 1 \cdot 2 = 2

= \frac{2}{5 5}

55 = 5

55

Apply radical rule:

a a = a

55 = 5

= 5

= \frac{2}{5}

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

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

(\frac{1}{5})^{2} = \frac{1}{5}

(\frac{1}{5})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{( 5 ) ^{2}}

(5)^{2} : 5

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 5

= \frac{1 ^{2}}{5}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{5}

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

(2 \cdot \frac{1}{5})^{2} = \frac{4}{5}

(2 \cdot \frac{1}{5})^{2}

Multiply

2 \cdot \frac{1}{5} : \frac{2}{5}

2 \cdot \frac{1}{5}

Multiply fractions:

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

= \frac{1 \cdot 2}{5}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{5}

= (\frac{2}{5})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{2 ^{2}}{( 5 ) ^{2}}

(5)^{2} : 5

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 5

= \frac{2 ^{2}}{5}

2^{2} = 4

= \frac{4}{5}

= (- \frac{1}{5} + 1) (- \frac{4}{5} + 1)

Join

1 - \frac{1}{5} : \frac{4}{5}

1 - \frac{1}{5}

Convert element to fraction:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{1}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 5 - 1}{5}

1 \cdot 5 - 1 = 4

1 \cdot 5 - 1

Multiply the numbers:

1 \cdot 5 = 5

= 5 - 1

Subtract the numbers:

5 - 1 = 4

= 4

= \frac{4}{5}

= \frac{4}{5} (- \frac{4}{5} + 1)

Join

1 - \frac{4}{5} : \frac{1}{5}

1 - \frac{4}{5}

Convert element to fraction:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{4}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 5 - 4}{5}

1 \cdot 5 - 4 = 1

1 \cdot 5 - 4

Multiply the numbers:

1 \cdot 5 = 5

= 5 - 4

Subtract the numbers:

5 - 4 = 1

= 1

= \frac{1}{5}

= \frac{4}{5} \cdot \frac{1}{5}

Multiply

\frac{4}{5} \cdot \frac{1}{5} : \frac{4}{25}

\frac{4}{5} \cdot \frac{1}{5}

Multiply fractions:

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

= \frac{4 \cdot 1}{5 \cdot 5}

Multiply the numbers:

4 \cdot 1 = 4

= \frac{4}{5 \cdot 5}

Multiply the numbers:

5 \cdot 5 = 25

= \frac{4}{25}

= \frac{4}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{4}{5}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{5}

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

Add similar elements:

\frac{2}{5} - \frac{2}{5} = 0

= 0

0 = 0

True

Plug in

x = - \frac{1}{5} :

True

(- \frac{1}{5}) \cdot 2 (- \frac{1}{5}) - (1 - (- \frac{1}{5})^{2}) (1 - (2 (- \frac{1}{5}))^{2}) = 0

(- \frac{1}{5}) \cdot 2 (- \frac{1}{5}) - (1 - (- \frac{1}{5})^{2}) (1 - (2 (- \frac{1}{5}))^{2}) = 0

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

Remove parentheses:

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

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

\frac{1}{5} \cdot 2 \cdot \frac{1}{5} = \frac{2}{5}

\frac{1}{5} \cdot 2 \cdot \frac{1}{5}

Multiply fractions:

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

= \frac{1 \cdot 1 \cdot 2}{5 5}

Multiply the numbers:

1 \cdot 1 \cdot 2 = 2

= \frac{2}{5 5}

55 = 5

55

Apply radical rule:

a a = a

55 = 5

= 5

= \frac{2}{5}

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

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

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

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

Apply exponent rule:

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

n

is even

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

= (\frac{1}{5})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{( 5 ) ^{2}}

(5)^{2} : 5

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 5

= \frac{1 ^{2}}{5}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{5}

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

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

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

Multiply

- 2 \cdot \frac{1}{5} : - \frac{2}{5}

- 2 \cdot \frac{1}{5}

Multiply fractions:

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

= - \frac{1 \cdot 2}{5}

Multiply the numbers:

1 \cdot 2 = 2

= - \frac{2}{5}

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

Apply exponent rule:

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

n

is even

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

= (\frac{2}{5})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{2 ^{2}}{( 5 ) ^{2}}

(5)^{2} : 5

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 5

= \frac{2 ^{2}}{5}

2^{2} = 4

= \frac{4}{5}

= (- \frac{1}{5} + 1) (- \frac{4}{5} + 1)

Join

1 - \frac{1}{5} : \frac{4}{5}

1 - \frac{1}{5}

Convert element to fraction:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{1}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 5 - 1}{5}

1 \cdot 5 - 1 = 4

1 \cdot 5 - 1

Multiply the numbers:

1 \cdot 5 = 5

= 5 - 1

Subtract the numbers:

5 - 1 = 4

= 4

= \frac{4}{5}

= \frac{4}{5} (- \frac{4}{5} + 1)

Join

1 - \frac{4}{5} : \frac{1}{5}

1 - \frac{4}{5}

Convert element to fraction:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{4}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 5 - 4}{5}

1 \cdot 5 - 4 = 1

1 \cdot 5 - 4

Multiply the numbers:

1 \cdot 5 = 5

= 5 - 4

Subtract the numbers:

5 - 4 = 1

= 1

= \frac{1}{5}

= \frac{4}{5} \cdot \frac{1}{5}

Multiply

\frac{4}{5} \cdot \frac{1}{5} : \frac{4}{25}

\frac{4}{5} \cdot \frac{1}{5}

Multiply fractions:

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

= \frac{4 \cdot 1}{5 \cdot 5}

Multiply the numbers:

4 \cdot 1 = 4

= \frac{4}{5 \cdot 5}

Multiply the numbers:

5 \cdot 5 = 25

= \frac{4}{25}

= \frac{4}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{4}{5}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{5}

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

Add similar elements:

\frac{2}{5} - \frac{2}{5} = 0

= 0

0 = 0

True

The solutions are

x = \frac{1}{5}, x = - \frac{1}{5}

x = \frac{1}{5}, x = - \frac{1}{5}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arccos (x) + arccos (2 x) = \frac{π}{2}

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

Check the solution

\frac{1}{5} :

True

\frac{1}{5}

Plug in

n = 1

\frac{1}{5}

For

arccos (x) + arccos (2 x) = \frac{π}{2}

plug in

x = \frac{1}{5}

arccos (\frac{1}{5}) + arccos (2 \cdot \frac{1}{5}) = \frac{π}{2}

Refine

1.57079 \dots = 1.57079 \dots

\Rightarrow True

Check the solution

- \frac{1}{5} :

False

- \frac{1}{5}

Plug in

n = 1

- \frac{1}{5}

For

arccos (x) + arccos (2 x) = \frac{π}{2}

plug in

x = - \frac{1}{5}

arccos (- \frac{1}{5}) + arccos (2 (- \frac{1}{5})) = \frac{π}{2}

Refine

4.71238 \dots = 1.57079 \dots

\Rightarrow False

x = \frac{1}{5}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for arccos(x)+arccos(2x)= pi/2 ?
The general solution for arccos(x)+arccos(2x)= pi/2 is x= 1/(sqrt(5))