Finding the range of $p$ such that $p = 3 \cos^2 x + 4 \sin x$

Question

Find the range of possible values for $p$ such that $$3 \cos^2 x + 4 \sin x = p$$

I tried: $$\begin{align} p &= \frac{1}{2}(8\sin(x) + 3\cos 2 x + 3) \tag1\\[4pt] &= -\frac32\sin^2x + 4\sin x + \frac32\cos^2x + \frac32 \tag2 \end{align}$$

Answer 1

We have $$3(1-\sin^2x)+4\sin{x}=p$$ or $$\sin^2x-\frac{4}{3}\sin{x}=1-\frac{p}{3}$$ or $$\left(\sin{x}-\frac{2}{3}\right)^2=\frac{13-3p}{9},$$ which gives firstly $$p\leq\frac{13}{3}.$$ The equality occurs for $\sin{x}=\frac{2}{3},$ which says that we got a maximal value.

Also, $p(\sin{x})=-3\sin^2x+4\sin{x}+3$ is a concave function, which says that $p$ gets a minimal value for an extreme value of $\sin,$ which happens for $\sin{x}=1$ or for $\sin{x}=-1.$

$p(1)=4$ and $p(-1)=-4$ and since $p$ is a continuous function we got the answer: $$\left[-4,\frac{13}{3}\right]$$

Answer 2

$$p=-3\sin^2x+4\sin x+3$$ and $\sin x\in[-1,1]$.

The quadratic function $-3s^2+4s+3$ achieves the maximum $$\dfrac{13}3$$ at $s=\dfrac23$, which is in the allowed range, and the minimum is reached at an endpoint of the interval, hence the smallest of

$$\pm4.$$