Minimum of $\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$

As $-1\leq\sin x\leq1\implies |\sin x-1|=1-\sin x$, $|\sin x-2|=2-\sin x$, $|\sin x-3|=3-\sin x$, $|\sin x+1|=1+\sin x$

Thus the given expression becomes, $1-\sin x+2-\sin x+3-\sin x+1+\sin x=7-2\sin x$ which is minimum when $\sin x $ is maximum and equal to $1$ which gives minimum value equal to $5.$

The number $\sin x$ is trying to decide where to live on the number line.

Her friends live at $-1$, $1$, $2$, and $3$.

She wants the sum of her distances to her friends to be as small as possible.

Where should $\sin x$ live? Unfortunately, she has to live between $-1$ and $1$. She starts at $-1$ and begins to travel to the right. For every tiny step she takes, her distance from $-1$ increases by $1$ step, and the sum of her distances from $1$, $2$, and $3$ decreases by $3$ steps, for a net decrease of $2$ steps.

So she should go to the right as far as her nature allows. I hope that $1$ doesn't mind when $\sin x$ moves in.

$\sin x-1$ is always non-positive.

$\sin x - 2$ is always negative.

$\sin x -3$ is always negative.

$\sin x + 1$ is always non-negative.

Therefore the whole expression reduces to $$ -(\sin x-1) -(\sin x-2) -(\sin x-3) + (\sin x+1). $$

(But if you meant $\sin(x-1)$ where you wrote $\sin x-1$, then none of the above will help you.)

You get $7-2\sin x$. Since the sine oscillates between $1$ and $-1$, the smallest this function ever gets is $7-2=5$.