I don't understand your description of the second solution of the second question, but your first solution of that question is correct, the range is $[1,\sqrt{2}]$.
The method for solving the first question is to follow definitions and think logically.
You know that $|\sin(x)| \in [0,1]$ and that $|\cos(x)| \in [0,1]$. Therefore $$|\sin(x)| + |\cos(x)| \in [0,2]
$$
It follows that
$$[|\sin(x)| + |\cos(x)|] \in \{0,1,2\}
$$
In other words, the range of your function is a subset of $\{0,1,2\}$.
So now you have three further questions to pursue:
- Does the range contain $0$?
- Does the range contain $1$?
- Does the range contain $2$?
Question 3 is the easiest: $[|\sin(x)| + |\cos(x)|] = 2$ if and only if $|\sin(x)| = 1$ and $|\cos(x)| = 1$, and I'm sure that you can convince yourself that this is impossible, no matter value of $x$ you pick. Therefore $2$ is not in the range of the function.
Question 2 is also easy: I'm sure that you can find a value of $x$ such that one of $|\sin(x)|$, $|\cos(x)|$ equals $0$ and the other equals $1$, so their sum equals $1$. Therefore $1$ is in the range of the function.
Question 1 is the trickiest. $[|\sin(x)| + |\cos(x)|] = 0$ if and only if $[|\sin(x)| + |\cos(x)|] \in [0,1)$. Can you find a value of $x$ for which this is true? Or perhaps can you prove that this is false for all values of $x$?
Perhaps that's enough for you to proceed.
Best Answer
Notice that for$ x=k\in \mathbb {Z}$ , we have $x-[x]=0$.
On the other hand if $k<x<k+1$ then $0<x-[x]=x-k<1$
Therefore the range of $x-[x]$ is $[0,1)$