For the range of a trigonometric biquadratic equation in sine where there is no cubic term, can we substitute f(0), f(1) and f(-b/2a) and find range?

Question

If we have a trigonometric biquadratic equation $ y = -4sin{^4}x + 6sin{^2}x + 2$ can we substitute the maximum and minimum value of $sin{^2}x$, $f(0)$ and $f(1)$ as well as $-b/2a$ (if it lies in the range $[0,1]$) to find the range of y?

Answer 1

Note maximizing $-4 x^2 + 6 x + 2$ over $\mathbb{R}$ is the same as maximizing $-4 g(x)^2 + 6 g(x) + 2$ over the range of $g(x)$.

Here $g(x) = sin^2 (x)$.

This holds for any $f(x) = h(g(x))$: maximizing $f(x)$ over a set $A$ is equivalent to maximizing $h(x)$ over $g(A)$.

Since the range of $g(x)$ is the closed interval $[0,1]$, the Closed Interval Theorem implies that we maximize $y = f(x)$ as you said.