$\cos{y}=x+\frac{1}{x}$, possible for any values of $y$? If for real values of x, $\cos{y}=x+\frac{1}{x}$,
Then how do we prove that no value of $y$ is possible ?

The value of $\cos(y)$ is between $-1$ and $1$.  Therefore, for this equality to hold, $x+\frac{1}{x}$ must be between $-1$ and $1$.


*

*If $x>1$, then $x+\frac{1}{x}>x>1$.

*If $x=1$, then $x+\frac{1}{x}=2>1$.

*If $0<x<1$, then $\frac{1}{x}>1$, so $x+\frac{1}{x}>\frac{1}{x}>1$.

*If $x=0$, then $x+\frac{1}{x}$ is not defined.

*If $-1<x<0$, then $\frac{1}{x}<-1$ so $x+\frac{1}{x}<\frac{1}{x}<-1$.

*If $x=-1$, then $x+\frac{1}{x}=-2<-1$.

*If $x<-1$, then $x+\frac{1}{x}<x<-1$.
Since $x+\frac{1}{x}$ is either greater than $1$ or less than $-1$, the equality can never hold (at least in the reals).
Alternately, one could use calculus and compute minima and limits as $x\rightarrow\pm\infty$ and $x\rightarrow 0^{\pm}$.  However, since the necessary inequality can be proved with only elementary facts, I chose this method.
