arcsin 1/2 + arcsin x = pi/3. Find the x value !

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...

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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`