Find range of $f(x)=3\cos^4x-6\cos^3x-6\cos^2x-3$ in the interval $[-π/2, π/2]$ While solving some questions related to functions, I came across this question and I am unable to find  its range.
$$f(x)=3\cos^4x-6\cos^3x-6\cos^2x-3$$in the interval $[-π/2, π/2]$
I tried graphing this function on Desmos and got this result.

However, since I can't use a graphing tool in exams, I need to be able to solve this algebraically.
When I tried factorising the function, I could only get this far:
$$f(x)=3(\cos x+1)(\cos^3x-3\cos^2x+\cos x-1)$$
Can anyone help how to solve this further?
 A: Here is a solution which does not use differentiation.
Let $$c=\cos \theta \in [0,1] \ \text{because} \ \theta \in [-\pi/2,\pi/2]$$
The given expression is
$$\varphi(c)=3c^4-6c^3-6c^2-3$$
Let us first observe that $\varphi(0)=-3$ and $\varphi(1)=-12$.
Therefore, according to the intermediate value theorem,
the range of $\varphi$ contains interval $[-12,-3]$.
Besides, let us express $\varphi(c)$ under the form:
$$\varphi(c)=3c^2((c-1)^2-3)-3 \tag{1}$$
As $0\le c \le 1$ , we have:
$$\begin{cases}-3&\le&(c-1)^2-3 &\le& -2\\
 \ \ \ 0 &\le& \ \ \ \ \ \ \ 3c^2& \le & \ \ \ 3\end{cases}$$
implying:
$$-9 \le 3c^2((c-1)^2-3) \le 0$$
A final subtraction gives:
$$-12 \le \underbrace{3c^2((c-1)^2-3)-3}_{\varphi(c)} \le -3$$
showing that the range of $\varphi$ is now contained into interval $[-12,-3]$.
As a consequence, the range of $f$ is precisely $[-12,-3]$.
A: Let $\varphi(x)=3x^4-6x^3-6x^2-3$; then $f(x)=\varphi\bigl(\cos(x)\bigr)$.
If $x\in[0,1]$, you have $\varphi'(x)\leqslant0$ and $\varphi'(x)=0\iff x=0$. So, $\varphi$ is decreasing on $[0,1]$. Since $\varphi(0)=-3$ and $\varphi(1)=-12$, the range of the restriction of $\varphi$ to $[0,1]$ is $[-12,-3]$ (see the graph of $\varphi$ below). Since the range of $\cos$ on $\left[-\frac\pi2,\frac\pi2\right]$ is $[0,1]$, the range of $f(=\varphi\circ\cos)$ is also $[-12,-3]$.

