Resolving $\sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$ Going steadily through my book, I found this exercise to resolve

$$ \sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$$

Here's how I resolve it ($LHS$) and again bear with me as I truly reverting to a feeling of vulnerability, like a child actually
As $\sec x$ is equal to $\frac{1}{\cos x}$
That leads us to this
$$\frac{(\sin^3x+\sin x\cos^2x)}{\cos x}$$
I'm factorizing one $\sin x$
$$\frac{\sin x(\sin^2x+\cos^2x)}{\cos x} = \frac{\sin x(1)}{\cos x} = \tan x$$
That seems to work otherwise I completly messed this up
Reading the book's solution, I have something different...
$$\begin{align*}
LHS&=\frac{\sin^3x}{\cos x}+ \sin x \cos x \\[4pt]
&=\frac{\sin x}{\cos x}-\frac{\sin x\cos^2x}{\cos x}+\sin x\cos x\\[4pt]
&= \tan x\end{align*}$$
What did I miss?

You missed nothing. The book is doing the same you did, only in a more cumbersome way.
