How do you solve #cos^3 (x) = cos (x)# on the interval #[0,2pi)#?

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Answer 1 · 2018-03-21T01:58:58

The only solutions on the given interval are #x=0, quad pi/2, quad pi, quad (3pi)/2#.

You can use a substitution, then solve it like a regular polynomial.

In this case, we'll substitute in #u# for #cosx#:

#color(white)=>cos^3x=cosx#

#color(white)=>(cosx)^3=cosx#

#=>u^3=u#

#color(white)=>u^3-u=0#

#color(white)=>u(u^2-1)=0#

#color(white)=>u(u-1)(u+1)=0#

#=>u=0,1,-1#

Now, put #cosx# back in for #u#:

#color(white)=>u=0,1,-1#

#=>cosx=0,1,-1#

Here's a unit circle to remind us of some cosine values:

#color(white)=>color(white){color(black)( (cosx=0,qquad cosx=1,qquad cosx=-1), (x=pi/2","(3pi)/2, qquad x=0, qquad x=pi)):}#

So the final answers are:

#x=0, quad pi/2, quad pi, quad (3pi)/2#

That's it. Hope this helped!