How do you draw the graph of #y=2-cosx# for #0<=x<2pi#?

This excercise involves function transformations. It means that you start from a function whose graph is known, and tranform it. Let's see what the transformations are and how they affect the graph:

Transformation 1: sign change

As a first step, we change from #cos(x)# to #-cos(x)#. In general, everytime you change from #f(x)# to #-f(x)# you reflect the graph vertically, with respect to the #x# axis. Here is the transformation:

Original function #f(x)=cos(x)#
graph{cos(x) [-0.2,6.48,-1.2,3.2]}

Reflected function #-f(x)=-cos(x)#
graph{-cos(x) [-0.2,6.48,-1.2,3.2]}

Transformation 2: vertical shift

The next transformation is represented by an additive constant: you change from #-cos(x)# to #-cos(x)+2#. In general, everytime you change from #f(x)# to #f(x)+k# you have a vertical translation, #k# units up if #k>0#, down otherwise. In this case, we have a translation of #2# units up. Here is the transformation:

Original function #f(x)=-cos(x)#
graph{-cos(x) [-0.2,6.48,-1.2,3.2]}

Shifted function #-f(x)=-cos(x)+2#
graph{-cos(x)+2 [-0.2,6.48,-1.2,3.2]}