$f(x)=sin(2x)-2sin(x)($Hint: This function has period $2,$ so you just need to

understand the graph on $0,2.)$

Solution: Zeroes: $x=0+2k,+2k$

Local Maxes: $(\frac{4}{3}+2k,\frac{3\sqrt[\phantom{2}]{3}}{2})$

Local Mins: $(\frac{2}{3}+2k,-\frac{3\sqrt[\phantom{2}]{3}}{2})$

Inflection Points:

$2k,1\mathrm{.}452$

Can you show me how they got the solution?