Question

Find the range of the function f(x) = (sin² x + sin x - 1)/(sin² x - sin x + 2).

Answer 1

Hence we get to the final answer.

Answer 2

\(f(x) = \frac{sin^2x + sinx -1}{sin^2x - sinx + 2}\)

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

\(\frac{2 ±\sqrt{4+4\times5\times7}}{2\times 7}=\frac{2\pm\sqrt{144}}{2\times 7}\)

\(y ∈ (\frac{3-2 \sqrt{11}}{7},\,\frac{3+2\sqrt{11}}{7})\)