Question

The range of $f\left(x\right)=\sqrt{{\sin }^{2}x-6\sin x+9}+3$ is [$-\infty$,a] Find a

Answer 1

$f\left(x\right)=\sqrt{{\sin }^{2}x-6\sin x+9}+3$
$=\sqrt{{(\sin x-3)}^2}+3$
$=|\sin x-3|+3$
Now $-1\le \sin x\le 1$
or $-4\le \sin x-3\le -2$
or $2\le |\sin x-3|\le 4$
or $5\le |\sin x-3|+3\le 7$
Ans: 7