Question

Find the range of $y={\sin }^{2}x+3\sin x+5$

Answer 1

Here $\sin x$ is bounded, that’s why we can not proceed .

$y={(\sin x+\frac{3}{2})}^2+5-\frac{9}{4}$

$\Rightarrow y={(\sin x+\frac{3}{2})}^2+\frac{11}{4}$

Since $-1\le \sin x\le 1$

$\Rightarrow \frac{1}{2}\le \sin x+\frac{3}{2}\le \frac{5}{2}$

$\Rightarrow \frac{1}{4}\le {(\sin x+\frac{3}{2})}^2\le \frac{25}{4}$

$\Rightarrow \frac{1}{4}+\frac{11}{4}\le {(\sin x+\frac{3}{2})}^2+\frac{11}{4}\le \frac{25}{4}+\frac{11}{4}$

$\Rightarrow \frac{12}{4}\le {(\sin x+\frac{3}{2})}^2\le \frac{36}{4}$

$\Rightarrow 3\le {(\sin x+\frac{3}{2})}^2\le 9$

Hence range of $f\left(x\right)=[3,9]$