Find the range of a function How to find the range of the following trigonometric function $\sin^2x-5\sin x-6$. Can some one help me out. Thank you

$$\sin^2x-5\sin x-6=\frac{(2\sin x)^2-2\cdot2\sin x\cdot5+5^2-24-25}4=\frac{(2\sin x-5)^2-49}4$$
Now, for real $\displaystyle x,-1\le\sin x\le1\iff -2\le\sin x\le2\iff-7\le2\sin x-5\le-3$
$\displaystyle2\sin x-5\le-3\implies(2\sin x-5)^2\ge9$
$\displaystyle2\sin x-5\ge-7\implies(2\sin x-5)^2\le49$
Can you take it home from here?
