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1 solution
f(x)=3|sin(x)| -2|cos(x)| Here we need to maximize the first term and minimize the second term.Max possible value of the 1st term =3 at x=pi/2 Similarly minimum value of second term will be 0 at x=pi/2 So max value=3
Similarly to minimize this we need max of second term and min of first which we get at x=0. i.e -2
Hence the range is[-2,3]