You should remember that `arcsin (a) = alpha` .

Calculating the sine function both sides yields:

`sin(arcsin 1/2 + arcsin x) = sin (pi/3)`

Use the trigonometric formula: `sin(alpha+beta) = sin alpha*cos beta + sin beta*cos alpha`

`sin(arcsin 1/2 + arcsin x) = sin (arcsin (1/2))*cos(arcsin x) + sin(arcsin x)*cos(arcsin (1/2))`

You need to remember that `sin (arcsin x) = x` and `cos(arcsin x) = sqrt(1 - x^2)`

`` `sin(arcsin 1/2 + arcsin x) = (1/2)*sqrt(1 - x^2) + x*sqrt(1 - 1/4)`

`` `sin(arcsin 1/2 + arcsin x) = (sqrt(1 - x^2))/2 + (xsqrt3)/2`

`` Write the equation:

`(sqrt(1 - x^2))/2 + (xsqrt3)/2 = sin (pi/3)`

`(sqrt(1 - x^2))/2 + (xsqrt3)/2 = sqrt3/2`

`` `sqrt(1 - x^2) + xsqrt3 = sqrt3`

`sqrt(1 - x^2) = sqrt3*(1-x)`

Raising to square both sides yields:

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

Expanding the binomial yields:

`1 - x^2 = 3 - 6x + 3x^2`

Bring all terms to the left side:

`1 - x^2- 3 +6x- 3x^2 = 0 =gt -4x^2 + 6x - 2 = 0`

Divide by -2:

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

Using quadratic formula yields:

`x_(1,2) = (3+-sqrt(9 - 8))/4 =gt x_(1,2) = (3+-1)/4`

`x_1 = 1 ; x_2 = 1/2`

**The solutions to the given equation are:`x_1 = 1 ; x_2 = 1/2` .**

Keep in mind that if arcsin x = y then sin y = x. To solve the problem first note that arcsin 1/2 = pi/6. Then subtract the pi/6 from both sides to get:

arcsin x = pi/3 - pi/6 = pi/6

And as we just saw arcsin 1/2 = pi/6 so x=1/2.