Question 381159
First of all when you set everything equal to zero you should obtain

{{{4 sin^3 (x) + 2 sin^2 (x) - 2 sin x - 1 = 0}}}

Let a = sin x. The equation then becomes

{{{4a^3 + 2a^2 - 2a - 1 = 0}}}

This is equivalent to {{{4a^3 - 2a + 2a^2 - 1 = 0}}}. We can factor {{{2a^2 -1}}} to obtain

{{{(2a^2 - 1)(2a + 1) = 0}}}

Solving, we obtain a = -1/2, sqrt(2)/2, and -sqrt(2)/2. All of these values are between -1 and 1, so we can find sin^-1 of each of these angles to find all values for x:

{{{sin^(-1) (-1/2) = (7pi/6)}}} or {{{(11pi/6)}}}


{{{sin^(-1) (sqrt(2)/2) = pi/4}}} or {{{3pi/4}}}


{{{sin^(-1) (-sqrt(2)/2) = 5pi/4}}} or {{{7pi/4}}}