Range of the trigonometric function $f(x)= \frac{3}{4 - 5\sin(x)}$ I was trying to find the range of the function
$$f(x) = \frac{3}{4-5\sin(x)} $$
The first thought that came to mind was:
Let's put the minimum and maximum value of $\sin(x)$ as $\{1,-1\}$ accordingly to find maximum and minimum of the function.
However, then I got range $[-3,\frac{1}{3}]$, which was totally wrong when I used a graphing calculator.
The true answer is $(-\infty,-3] ∪ [\frac{1}{3}, \infty)  $

Now, I want explaination what I did wrong here.

Link for graph
 A: Let the denominator be $y$. Then, as you observed $-1\leq y\leq 9$. But you have to be careful when find the range for $\frac  1 y$. At $y=0$, $\frac  1 y$ is not defined. You have to consider $0<y\leq 9$ and $-1\leq y<0$ separately. For $0<y\leq 9$ we get $\frac  1 y \in [\frac 1 9, \infty)$ and for $-1\leq y<0$ we get $\frac  1 y \in (-\infty, -1]$. Now multiply by $3$.
A: To really understand how to avoid arriving at an incorrect conclusion, it is important to consider not just the extrema of the interval that $\sin x$ maps to, but also the intermediate values that are possible.  To do this, observe
$$-1 \le \sin x \le 1 \tag{1}$$ implies $$5 \ge -5 \sin x \ge -5. \tag{2}$$  (Recall that multiplying an inequality by a negative number reverses the direction of the inequality.)  Hence $$4 + 5 \ge 4 - 5 \sin x \ge 4 - 5$$ or equivalently $$9 \ge 4 - 5 \sin x \ge -1. \tag{3}$$  Now, here is where the mistake is made:  if you took reciprocals of this inequality, you might think it will look like this:
$$\frac{1}{9} \le \frac{1}{4 - 5 \sin x} \le -1.$$  Again, remember that taking reciprocals, like negation, will reverse the direction of the inequality.  But this inequality does not make any sense because $1/9 > -1$.  In the previous step $(3)$, clearly the choice $x = 0$ would give us a valid result, $9 \ge 4 \ge -1$.  But after taking reciprocals, $1/9 \le 1/4$ is still true, but $1/4 \le -1$ is false.  A piece of reasoning is missing.
The missing piece is that the inequality $(3)$ contains $0$.  If I had written $$9 \ge y \ge -1,$$ then $y = 0$ is a valid solution to this inequality; however, upon taking reciprocals, $1/y$ is not defined when $y = 0$.  So this is the first hint.  The second hint is that if $y$ is negative, say $y = -1/2$, then taking reciprocals gives us $9 \le -2 \le -1$ which is true on the right-hand side but not the left.  So what we need to do here is split the compound inequality $(3)$ into separate inequalities according to whether the middle term is positive or negative (zero is not allowed):
$$9 \ge 4 - 5 \sin x > 0, \quad {\bf\text{or}} \quad 0 > 4 - 5 \sin x \ge -1. \tag{4}$$  Now that we have done this, taking reciprocals of each will no longer result in a contradiction:
$$\frac{1}{9} \le \frac{1}{4 - 5 \sin x} < \infty, \quad \text{or} \quad -\infty < \frac{1}{4 - 5 \sin x} \le -1. \tag{5}$$  Now we multiply by $3$ and drop the infinities (because they are implied) to get
$$\frac{1}{3} \le \frac{3}{4 - 5 \sin x}, \quad \text{or} \quad \frac{3}{4 - 5 \sin x} \le -3. \tag{6}$$  This gives us the desired range of the function.
A: Although somewhat convoluted, a nice approach to finding range could be by finding the domain of the inverse function.
Notice that if $f(x)={{3}\over{4-5\sin{x}}}$ then $f^{-1}(x)=\sin^{-1}\big({4\over5}-{3\over{5x}}\big)$ making the domain of $f^{-1}(x)$ satisfying:
$$-1\leq{4\over5}-{3\over{5x}}\leq1$$
which simplifies to:
$$3\geq{1\over x}\geq-{1\over3}$$
Looking at where equality holds, we get $x={1\over3}$ or $x=-3$. Testing for when the inequality above holds by using test values around $x={1\over3}$ and $x=-3$ on the entire real line (say $x=-4,\ -1,\ 1$) we get that $x\in(-\infty,-3]\cup\big[{1\over3},\infty\big)$, which is also the range of $f(x)$, as desired.
A: Instead, look at the range of the function $$g(x)=\frac{1}{f(x)}=\frac{4-5\sin(x)}{3}$$
What is the range of $g(x)$?
$$-1\le\sin(x)\le1\Rightarrow\frac{-1}{3}\le\frac{4-5\sin(x)}{3}\le3$$
Hence, range of $g(x)$ is $\left[\frac{-1}{3},3\right]$. What should be the range of the inverse of this?
$$(-\infty,-3]\cup[\frac{1}{3},\infty)$$
Note that for $f(x)$ to exist, $\sin(x)\ne\frac{4}{5}$. Make sure the domain doesn't include this particular $x$.
