Find the range of $f (x) = 3 {sin}^{2} x + sin 2 x + 3$

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Solution

$f (x) = 3 {sin}^{2} x + sin 2 x + 3$
$= 3 (\frac{1 - cos 2 x}{2}) + sin 2 x + 3$
$= sin 2 x - \frac{3}{2} cos 2 x + \frac{9}{2}$
$= \sqrt{1^{2} + \frac{3^{2}}{2^{2}}} sin (2 x - ϕ) + \frac{9}{2}$
Hence,Range $\in [\frac{9 - \sqrt{13}}{2}, \frac{9 + \sqrt{13}}{2}]$

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