$f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable? $f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable?

1.$\{(2n+1){\pi\over2}\}$

2.$\{n\pi\}$

3.$\{{n\pi\over 2}\}$

4.$\{n\pi+2\}$

in all cases $n\in\mathbb{Z}$

well, is there any easy trick to solve this type of probelem in an competitive examination? Do I need to pick $\pi \over 2$ from $1$ and cheked that $f$ is differentiable and same trick from each set? in some what I am lost.
A function isn't differentiable where it has sharp corners since the tangent line at that point is not well-defined. In this case, it fails to be differentiable when $\cos(x)$ and $\sin(2-x)$ change sign since the absolute value of a function has a sharp cusp when its argument changes sign.
Hint: The function $ g(y)=|y|$ isn't differentiable at $ y=0$, and therefore, $ g(\cos x)+g(\sin(2-x))$ isn't differentiable when it's arguments are $0$, meaning, when $ \cos x=0$ or when $ \sin (2-x)=0$