[Algebra]
Problem: If $\frac{1}{\cos x}-\tan x=3$, what is the numerical value of $\sin x$ ?

Solution: Beginning with the given equation, we have

$$
\begin{aligned}
\frac{1}{\cos x}-\tan x & =3 \\
\frac{1}{\cos x}-\frac{\sin x}{\cos x} & =3 \\
1-\sin x & =3 \cos x \quad(\text { since } \cos x \neq 0) \\
(1-\sin x)^{2} & =9 \cos ^{2} x \quad \text { (squaring both sides) } \\
1-2 \sin x+\sin ^{2} x & =9\left(1-\sin ^{2} x\right) \\
10 \sin ^{2} x-2 \sin x-8 & =0 \\
5 \sin ^{2} x-\sin x-4 & =0 \\
(5 \sin x+4)(\sin x-1) & =0
\end{aligned}
$$

Therefore, $\sin x=-\frac{4}{5}$ or $\sin x=1$.

If $\sin x=1$, then $\cos x=0$ and $\tan x$ is undefined, which is inadmissible in the original equation.

Therefore, $\sin x=-\frac{4}{5}$.

(We can check that if $\sin x=-\frac{4}{5}$, then $\cos x= \pm \frac{3}{5}$ and the possibility that $\cos x=\frac{3}{5}$ satisfies the original equation, since in this case $\frac{1}{\cos x}=\frac{5}{3}$ and $\tan x=-\frac{4}{3}$ and the difference between these fractions is 3 .)

Final answer: $-\frac{4}{5}$