\(f(x) = \frac{sin^2x + sinx -1}{sin^2x - sinx + 2}\)
sin x → t
\(\frac{t^ 2+t-1}{t^ 2-t+2}=y\)
t2 + t - 1 = yt2 - yt + 2y
(y-1)t2 - t(y+1) + 2y + 1 = 0
D ≥ 0
b2 - 4ac ≥ 0
(y+1)2 - 4(y-1) (2y+1) ≥ 0
y2 + 1 - 2y - (4(2y2 + y - 2y - 1)) ≥ 0
y2 + 1 - 2y - 8y2 + 4y + 4 ≥ 0
-7y2 + 2y + 5 ≥ 0
7y2 - 2y - 5 ≤ 0
D ≥ 0
7y2 - 2y - 5 = 0
\(\frac{2 ±\sqrt{4+4\times5\times7}}{2\times 7}=\frac{2\pm\sqrt{144}}{2\times 7}\)
\(y ∈ (\frac{3-2 \sqrt{11}}{7},\,\frac{3+2\sqrt{11}}{7})\)