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!

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