What is the value set of the function $h(x) = - \frac{4}{5} \sin{\left(\frac{5\pi x}{6} \right )} - 1$

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The assignment:

Let us start by defining $f:\mathbb{R} \to ]-\infty,2]$ as $f(x) = - \frac{4}{5} \sin{\left (\pi x \right )} - 1$, and $g:\mathbb{R}\to\mathbb{R}$ as $g(x) = \frac{5 x}{6}$. In this assignment we are going to study the composite function $h$ of $f$ and $g$, which fulfills $h(x)=f(g(x))$ for all $x$ in its definition set.

I am stuck with the part to determine the value set for the composite function $h(x)=f(g(x)) = - \frac{4}{5} \sin{\left(\frac{5\pi x}{6} \right )} - 1$

I have already determined the definition set and goal set for the function:

$g$ determines the definition set and $f$ determines the goal set. Because $g:\mathbb{R}\to\mathbb{R}$ and $f:\mathbb{R} \to ]-\infty,2]$ we get that: $h:\mathbb{R}\to\mathbb{R}\to ]-\infty, 2]$ which gives us $h:\mathbb{R}\to ]-\infty, 2]$

Now I should determine the value set of the function and I am unsure of how to think. I understand that I may have to do some test for some x values and see a pattern and from that determine the value set. I know that sine has the value set of $[-1,1]$. According to the definition set of the function $h$ I should be able to enter whatever real number x and get a result according to the goal set of $[-\infty, 2]$ . But from here I am getting unsure of which values of x to test for. Which real numbers can I test for?

When I do some tests for $x = -1$, $x = 0$ and $x = 1$ I get $-\frac{3}{5}$, $-1$ and $- \frac{7}{5}$. But I came to the conclusion that this is not enough. Because when I test further for $x = 2$, $x = 3$, $x = 4$ and $x = 5$ I get $\frac{2 \sqrt{3}}{5} - 1$, $-\frac{9}{5}$, $\frac{2 \sqrt{3}}{5} - 1$ and $- \frac{7}{5}$. When I test for some more negative numbers, $x = -2$, $x = -3$ and $x = -4$ I get $-\frac{2 \sqrt{3}}{5} - 1$, $- \frac{1}{5}$ and $-\frac{2 \sqrt{3}}{5} - 1$.

From that pattern I can see that the value set for the function is $- \frac{1}{5}$, $-\frac{2 \sqrt{3}}{5} - 1$, $-\frac{3}{5}$, $-1$, $- \frac{7}{5}$, $\frac{2 \sqrt{3}}{5} - 1$, $-\frac{9}{5}$. This is because of the periodicity of sine. But I am unsure of my answer. Is this correct? I am also thinking that real numbers doesnt have to only be integers. How should I think? Is there another way do determine which x values to test for to get the value set for the function?