Calculate local maximum and minimum of $f(x)=(2+\sin x)(5-\sin x)$ I am working on my scholarship exam practice which assumes high school or pre-university math knowledge. Could you please have a look on my approach?

The minimum of the function $f(x)=(2+\sin x)(5-\sin x)$ is ......

First, I began with some basic approach. 
We know that $-1\leq\sin x\leq1$, so I just test both values $-1$ and $1$ in the function.
$f(-1)=(2+(-1))(5-(-1))=6$
$f(1)=(2+1)(5-1)=12$
Since $-1$ is the minimum of function $\sin x$, I conclude that $f(-1)=6$, which is correct when checked with the value of answer key provided. Please let me know if my approach is not always true or can apply on other similar problems.
But in exam I may be uncertain if my answer is right so I tried calculus approach to check my answer.
$f(x)=(2+\sin x)(5-\sin x)=10+3\sin x-\sin^2 x$
$f'(x)=3\cos x-2\sin x\cos x=0$
$2\sin x\cos x=3\cos x$
$2\sin  x=3$
$\sin  x=\frac{3}{2}=1.5>1$
Since 1.5 exceeds the range of $\sin x$, then I cannot use this approach. I am wondering why this is the case. Why can't we use this method to find the minimum?
 A: We have $2 \sin x \cos x=3\ cos x \iff 2 \sin x =3$ or $ \cos x=0.$
A: $f(x)=10+3\sin{x}-\sin^2x$ and since $-1<1<\frac{3}{2},$ 
we see that  $y=10+3t-t^2$ increses on $[-1,1]$.
Thus, $$6=f\left(-\frac{\pi}{2}\right)\leq f(x)\leq f\left(\frac{\pi}{2}\right)=12.$$
A: You cannot divide by $\cos x$ since it may be equal to zero.
It is true that
$$f(x) = (2 + \sin x)(5 - \sin x) = 10 + 3\sin x - \sin^2x$$
Therefore,
$$f'(x) = 3\cos x - 2\sin x\cos x$$
Setting the derivative equal to zero yields
\begin{align*}
f'(x) & = 0\\
3\cos x - 2\sin x\cos x & = 0\\
\cos x(3 - 2\sin x) & = 0
\end{align*}
Hence, 
\begin{align*}
\cos x & = 0 & 3 - 2\sin x & = 0\\
x & = \frac{\pi}{2} + n\pi, n \in \mathbb{Z} & -2\sin x & = -3\\
  &                                          & \sin x & = \frac{3}{2}
\end{align*}
The equation $\sin x = \frac{3}{2}$ has no real-valued solutions.  Notice that we can express 
$$x = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$$
as 
$$x = \begin{cases}
      \dfrac{\pi}{2} + 2k\pi, k \in \mathbb{Z}\\
      -\dfrac{\pi}{2} + 2m\pi, m \in \mathbb{Z}
      \end{cases}
$$
which will be useful since the original function has sines and the value sine takes at $\frac{\pi}{2}$ is different from the value it takes at $-\frac{\pi}{2}$.
Notice that since $3 - 2\sin x > 0$ for each $x \in \mathbb{R}$, the sign of the derivative is determined by $\cos x$, which is positive in the intervals $$(-\frac{\pi}{2} + 2n\pi, \frac{\pi}{2} + 2n\pi), n \in \mathbb{Z}$$ and negative in the interval $$(\frac{\pi}{2} + 2n\pi, \frac{3\pi}{2} + 2n\pi), n \in \mathbb{Z}$$  By the First Derivative Test, $f$ has relative maxima at $$x = \frac{\pi}{2} + 2n\pi$$ and relative minima at $$x = -\frac{\pi}{2} + 2\pi$$
If $x = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z}$, $\sin x = 1$.  Thus, the relative maximum value $f$ assumes is 
$$f(\frac{\pi}{2} + 2k\pi) = f\left(\frac{\pi}{2}\right) = (2 + 1)(5 - 1) = 3 \cdot 4 =12$$
If $x = -\frac{\pi}{2} + 2m\pi, m \in \mathbb{Z}$, $\sin x = -1$.  Thus, the relative minimum value $f$ assumes is 
$$f(-\frac{\pi}{2} + 2k\pi) = f\left(\frac{3\pi}{2}\right) = (2 - 1)(5 + 1) = 1 \cdot 6 = 6$$
which should look familiar.
A: $$f(x)=(2+\sin x)(5-\sin x) =  -\sin^2 x + 3\sin x + 10$$
 this function has $D<0$. And obviously minimum when $\sin x$ is $-1$ and max when $\sin x$ is $1$.
