(i) We start with the expression \(f(x) = 5 \sin^2 x + 3 \cos^2 x\). Using the identity \(\sin^2 x + \cos^2 x = 1\), we can rewrite \(\cos^2 x\) as \(1 - \sin^2 x\). Thus,
\(f(x) = 5 \sin^2 x + 3(1 - \sin^2 x) = 5 \sin^2 x + 3 - 3 \sin^2 x = 3 + 2 \sin^2 x.\)
Therefore, \(a = 3\) and \(b = 2\).
(ii) We need to solve \(3 + 2 \sin^2 x = 7 \sin x\). Let \(s = \sin x\), then the equation becomes:
\(3 + 2s^2 = 7s.\)
Rearranging gives:
\(2s^2 - 7s + 3 = 0.\)
Using the quadratic formula \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = -7, c = 3\), we find:
\(s = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}.\)
This gives \(s = 3\) or \(s = \frac{1}{2}\). Since \(s = \sin x\) and \(0 \leq x \leq \pi\), \(s = 3\) is not possible. Thus, \(s = \frac{1}{2}\), which gives \(x = \frac{\pi}{6}\) or \(x = \frac{5\pi}{6}\).
(iii) The minimum value of \(f(x) = 3 + 2 \sin^2 x\) occurs when \(\sin^2 x = 0\), giving \(f(x) = 3\). The maximum value occurs when \(\sin^2 x = 1\), giving \(f(x) = 5\). Thus, the range of \(f\) is \(3 \leq f(x) \leq 5\).
