# What is root in equation arcsinx-arccosx=arcsin(3x-2)?

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### 2 Answers

You should take sine function both sides, such that:

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

Expanding the left side yields:

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

You need to use the following trigonmetric identities, such that:

`sin(arcsin x) = cos(arccos x) = x`

`sin(arccos x) = cos(arcsin x) = sqrt(1 - x^2)`

Using the identities in the expansion above, yields:

`x*x - sqrt(1 - x^2)*sqrt(1 - x^2) = 3x - 2`

`x^2 - (1 - x^2) = 3x - 2`

`x^2 - 1 + x^2 - 3x + 2 = 0`

`2x^2 - 3x + 1 = 0`

Using quadratic formula yields:

`x_(1,2) = (3+-sqrt(9 - 8))/4 => x_(1,2) = (3+-1)/4 => x_1 = 1; x_2 = 1/2`

Testing the solutions into the given equation yields:

`x = 1 => arcsin 1 - arccos 1 = arcsin (3 - 2)`

`pi/2 - 0 = arcsin 1 = pi/2 => pi/2 = pi/2` valid

`x = 1/2 => arcsin (1/2) - arccos (1/2) = arcsin (3/2 - 2)`

`pi/6 - pi/3 = arcsin(-1/2) = -pi/6`

`(pi - 2pi)/6 = -pi/6`

`-pi/6 = -pi/6` valid

**Hence, evaluating the solutions to the given equation, under the given conditions, yields **`x = 1/2, x = 1.`

**Sources:**

Given question is

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

since

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

`arccos(x)=pi/2-arcsin(x)`

`therefore`

`arcsin(x)-(pi/2-arcsin(x))=arcsin(3x-2)`

`2arcsin(x)-pi/2=arcsin(3x-2)`

`sin(2arcsin(x))-pi/2)=3x-2`

`sin(2arcsin(x))cos(pi/2)-cos(2arcsin(x))sin(pi/2)=3x-2`

`-cos(2arcsin(x))=3x-2`

`cos(2arcsin(x))=2-3x`

let arcsin(x)=y

x=sin(y)

therefore

cos(2y)=2-3sin(y)

`1-2sin^2(y)=2-3sin(y)`

`2sin^2(y)-3sin(y)+1=0`

`(sin(y)-1)(2sin(y)-1)=0`

Either

sin(y)-1=0

sin(y)=1

`y=pi/2`

`arcsin(x)=pi/2`

`x=sin(pi/2)=1`

OR

2sin(y)-1=0

2sin(y)=1

sin(y)=1/2

`y=pi/6`

`arcsin(x)=pi/6`

`x=sin(pi/6)=1/2`

**Thus either x=1 or x=1/2.**