You may reduce the trigonometric equation to a quadratic equation performing the substitution `arcsin x = t` , such that:

`2t^2 - t - 6 = 0`

Using quadratic formula yields:

`t_(1,2) = (1+-sqrt(1 + 48))/4`

`t_(1,2) = (1+-sqrt 49)/4 => t_(1,2) = (1+-7)/4 => {(t_1 = 2),(t_2 = -3/2):}`

You need to solve for x the equations `arcsin x = t_(1,2)` , such that:

`arcsin x = 2`

Since `2 > 1.57 = pi/2` and since the range of the inverse trigonometric function `arcsin x in [-pi/2,pi/2]` yields that `arcsin x = 2` is an invalid statement.

`arcsin x = -3/2`

Since `-3/2 = -1.5 > -1.57 = -pi/2` yields that the equation `arcsin x = -3/2` can be solved, such that:

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

Convert `-3/2` in radians using the formula:

`alpha = (-3/2*pi)/3.14 => alpha ~~ (24pi)/5 rad`

`x = -sin ((24pi)/5) => x ~~ 0.008`

**Hence, evaluating the solution to the given equation, under the given conditions, yields `x ~~ 0.008` .**