What is the general solution for arcsin(x)-arccos(x)=arcsin(3x-2) ?

The general solution for arcsin(x)-arccos(x)=arcsin(3x-2) is x=1,x= 1/2

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arcsin(x)-arccos(x)=arcsin(3x-2)

Solution

arcsin (x) - arccos (x) = arcsin (3 x - 2)

Solution

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

Solution steps

arcsin (x) - arccos (x) = arcsin (3 x - 2)

a = b \Rightarrow sin (a) = sin (b)

sin (arcsin (x) - arccos (x)) = sin (arcsin (3 x - 2))

Use the following identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

sin (arcsin (x)) cos (arccos (x)) - cos (arcsin (x)) sin (arccos (x)) = sin (arcsin (3 x - 2))

Use the following identity:

sin (arcsin (x)) = x

Use the following identity:

cos (arccos (x)) = x

Use the following identity:

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

Use the following identity:

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

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

Solve

xx - 1 - x^{2} 1 - x^{2} = 3 x - 2 : x = 1, x = \frac{1}{2}

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

Expand

xx - 1 - x^{2} 1 - x^{2} : 2 x^{2} - 1

xx - 1 - x^{2} 1 - x^{2}

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}

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

1 - x^{2} 1 - x^{2}

Apply radical rule:

a a = a

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

= 1 - x^{2}

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

Expand

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

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

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

- (1 - x^{2})

Distribute parentheses

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

Apply minus-plus rules

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

= - 1 + x^{2}

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

= 2 x^{2} - 1

= 2 x^{2} - 1

= 2 x^{2} - 1

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

Solve

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

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

Move

2

to the left side

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

Add

2

to both sides

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

Simplify

2 x^{2} + 1 = 3 x

2 x^{2} + 1 = 3 x

Move

3 x

to the left side

2 x^{2} + 1 = 3 x

Subtract

3 x

from both sides

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

Simplify

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

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

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

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

(- 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

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

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

Verify Solutions:

x = 1

True

, x = \frac{1}{2}

True

Check the solutions by plugging them into

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

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

Plug in

x = 1 :

True

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 \cdot 1 - 1 - 1 1 - 1

1 \cdot 1 = 1

1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 1

1 - 1 1 - 1 = 0

1 - 1 1 - 1

1 - 1 = 0

1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

Apply rule

0 = 0

= 0

= 0 \cdot 1 - 1

1 - 1 = 0

1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

Apply rule

0 = 0

= 0

= 0 \cdot 0

Multiply the numbers:

0 \cdot 0 = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

3 \cdot 1 - 2 = 1

3 \cdot 1 - 2

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 2

Subtract the numbers:

3 - 2 = 1

= 1

1 = 1

True

Plug in

x = \frac{1}{2} :

True

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

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

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

Remove parentheses:

(a) = a

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

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1}{4}

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

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

Apply radical rule:

a a = a

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

2^{2} = 4

= \frac{1}{4}

= 1 - \frac{1}{4}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 4 - 1 = 3

1 \cdot 4 - 1

Multiply the numbers:

1 \cdot 4 = 4

= 4 - 1

Subtract the numbers:

4 - 1 = 3

= 3

= \frac{3}{4}

= \frac{1}{4} - \frac{3}{4}

Apply rule

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

= \frac{1 - 3}{4}

Subtract the numbers:

1 - 3 = - 2

= \frac{- 2}{4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

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

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

Remove parentheses:

(a) = a

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

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

3 \cdot \frac{1}{2}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{2}

= \frac{3}{2} - 2

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

- 2 \cdot 2 + 3 = - 1

- 2 \cdot 2 + 3

Multiply the numbers:

2 \cdot 2 = 4

= - 4 + 3

Add/Subtract the numbers:

- 4 + 3 = - 1

= - 1

= \frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

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

True

The solutions are

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

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

arcsin (x) - arccos (x) = arcsin (3 x - 2)

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

Check the solution

1 :

True

1

Plug in

n = 1

1

For

arcsin (x) - arccos (x) = arcsin (3 x - 2)

plug in

x = 1

arcsin (1) - arccos (1) = arcsin (3 \cdot 1 - 2)

Refine

1.57079 \dots = 1.57079 \dots

\Rightarrow True

Check the solution

\frac{1}{2} :

True

\frac{1}{2}

Plug in

n = 1

\frac{1}{2}

For

arcsin (x) - arccos (x) = arcsin (3 x - 2)

plug in

x = \frac{1}{2}

arcsin (\frac{1}{2}) - arccos (\frac{1}{2}) = arcsin (3 \cdot \frac{1}{2} - 2)

Refine

- 0.52359 \dots = - 0.52359 \dots

\Rightarrow True

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

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Frequently Asked Questions (FAQ)

What is the general solution for arcsin(x)-arccos(x)=arcsin(3x-2) ?
The general solution for arcsin(x)-arccos(x)=arcsin(3x-2) is x=1,x= 1/2