Question 377998
Let a = 2 sin (2x). This expression becomes {{{2a^2 + 5a - 3 = 0}}}. The equation factors to (2a - 1)(a + 3) = 0, so we have a = 1/2 or a = 3. Therefore we have

2 sin (2x) = 1/2 --> sin (2x) = 1/4
2 sin (2x) = -3 --> sin (2x) = -3/2

The second equation has no solutions for x, as the sine function is defined with outputs [-1,1]. The first one has solution {{{2x = arcsin(1/4)}}} or

{{{x = (arcsin(1/4))/2}}}

Note that we may add any integer multiple of pi to x, as sine is periodic with period 2pi (note that x is multiplied by 2).