Find the derivative:
#2cosx-sinx#
#d/dx[2cosx-sinx]=2(-sinx)-cosx#
#=-2sinx-cosx#
The maximum and minimum is achieved when #f'(x) =0#
#0=-2sinx-cosx#
#cosx=-2sinx#
#-1/2=(sinx)/(cosx) #
#-1/2=tanx#
#tan^(-1)(-1/2)=x#
#x=-0.464...#
A more general solution is:
#x=-0.464+kpi#
Where #k# is an integer.
Since #tanx# has a period of #pi#.
However it does differentiate the values for max and min.
Let us try when #x=-0.464# that is when #k=0# in the original function. This would achieve the max value since #-sin(-0.464)=+sin(0.464)# and #cosx# will be positive.
When #x=-0.464+2pi# it would achieve the same thing.
However when #x=-0.464+pi# it would make the entire function would negative
Thus:
Max:#-0.464+(2k)pi#
Min:#-0.464+(2k+1)pi#
