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

sciencesolve | Certified Educator

You need to remember that the range of cosine function is `[-1,1]` , hence, you need to solve the following inequality such that:

`-1 <= (x+x^-1)/2 <= 1 => {((x + (1/x))/2 >= -1),((x + (1/x))/2 <= 1):}`

`(x^2 + 1)/(2x) >= -1 => (x^2 + 1)/(2x) + 1 >= 0`

`(x^2 + 2x + 1)/(2x) >= 0 => (x+1)^2/(2x) >= 0`

Since `(x+1)^2 >= 0` , hence `2x > 0 => x > 0.`

`(x + (1/x))/2 <= 1 => (x^2 - 2x + 1)/(2x) <= 0 => (x - 1)^2/(2x) <= 0`

`Since (x - 1)^2 >= 0 => 2x < 0 => x < 0`

You need to remember that the domain of cosine function is [0,pi], hence, you need to solve the following inequality such that:

`0 <= y <= pi`

Hence, evaluating the values of x and y, under the given conditions, yields `x in (-oo,0)U(0,oo)` and `y in [0,pi].`

giorgiana1976 | Student

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