# Solve for x and y. cos y=(x+x^-1)/2

*print*Print*list*Cite

### 1 Answer

We'll apply the inequality between the arithmetical mean and geometrical mean:

for a>0, b>0, then

(a+b)/2>sqrt (a*b)

Particularly, fora>0, then

a+(1/a)>2sqrt(a*1/a)=2, and the equality takes place if only a=1.

If a<0, then a+(1/a)<-2 and the equality takes place if only a=-1.

We know that: -2<2cos y<2

From all the above, the result is if x>0, then

x+(1/x)>2>2cos y (Note: we'll write x^-1 = 1/x)

We can have solutions if only

x+(1/x)=2 and

2 cos y=2 if x=1

and

y=2k*pi, where k is in Z set.

If x<0, then x+(1/x)<-2<2 cos y

We can have solutions if only x+(1/x)=-2 and 2 cos y=-2, meaning if x=-1 and y= pi + 2k*pi

The roots of the equation x+(1/x)=2 cos y are:

x=1 and y=2k*pi

x=-1 and y=pi+2k*pi, where k is integer