What is the asymptotic form of $\frac {3\cos x}x-\sqrt{(\frac {\cos x}x+\sin x-y)^2-1}$ in terms of $\mathcal{O}(\frac 1x)$ as $x\to\infty$?

Question

I have this equation for a variable $x\in(0,\infty)$ and a constant $y\in[-1,1]$

$$ f(x)=0 ,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\\ f(x)=\frac {3\cos x}x-\sqrt{\left(\frac {\cos x}x+\sin x-y\right)^2-1} $$ with $\frac {\cos x}x+\sin x-y>1$ to get real solutions.

My Question

How can I write the function $f(x)$ in the asymptotic form in terms of $\mathcal{O}(\frac 1x)$ as $x\to\infty$? And then what would be the necessary condition for having real solutions?

I know that $\sin x$ is indeterminate in the limit $x\to\infty$, but, are we allowed to say that as $x\to\infty$, the function $f(x)$ behaves asymptotically in a form such as this? $$ f(x)= -\sqrt{\left(\mathcal{O}(\frac 1x)+\sin x- y\right)^2-1}+\mathcal{O}(\frac 1x) $$

P.S. My major is not mathematics, so, please guide me if my question is wrong instead of downvoting:)

Answer 1

$$\frac{\cos(x)}{x}+\sin(x)-\frac{1}{2}\le\frac{1}{x}+1-\frac{1}{2}=\frac{1}{x}+\frac{1}{2}$$ $\frac{1}{x}+\frac{1}{2}$ (upper limit) becomes smaller than $1$ (lower limit) as soon as $x>2$, thus disallowing real solutions in that region.

i.e. question is malformed, sadly