2 sin(x) + 3 > sin²(x)
0 > sin²(x) - 2 sin(x) - 3
0 > (sin(x) - 3) (sin(x) + 1)
The right side is 0 when either
sin(x) - 3 = 0 or sin(x) + 1 = 0
sin(x) = 3 or sin(x) = -1
The first case offers no real solutions, so we're left with
sin(x) = -1
which happens for
x = -π/2 + 2nπ
where n is any integer.
In the interval 0 ≤ x ≤ 2π, the expression is 0 for x = 3π/2 (when n = 1). So check the sign of the expression when x is picked from two different intervals:
• If 0 ≤ x < 3π/2, then sin(x) - 3 < 0 and sin(x) + 1 > 0, so their product sin²(x) - 2 sin(x) - 3 < 0.
• If 3π/2 < x ≤ 2π, then sin(x) - 3 < 0 and sin(x) + 1 > 0, so the product is again negative.
In both cases, sin²(x) - 2 sin(x) - 3 < 0, so we have the solution set
0 ≤ x < 3π/2 and 3π/2 < x ≤ 2π
Another way to arrive at the same conclusion:
0 > sin²(x) - 2 sin(x) - 3
Recall the half-angle identity for sin:
sin²(x) = (1 - cos(2x))/2
So we have
0 > (1 - cos(2x))/2 - 2 sin(x) - 3
0 > -cos(2x) - 4 sin(x) - 7
0 < cos(2x) + 4 sin(x) + 7
Both sin(x) and cos(x) are bounded between -1 and 1. If they are both minimized, so that cos(2x) = sin(x) = -1, then the right side is at least
-1 + 4(-1) + 7 = 2
so the inequality holds for all x in the interval, except for x = 3π/2.