<s>[INST] Show that $\frac{3\> + \>\cos x}{\sin x}$ cannot have any value between $-2\sqrt2$ and $2\sqrt2$ Show that 
$$\dfrac{3+\cos x}{\sin x}\quad \forall \quad x\in R $$
cannot have any value between $-2\sqrt{2}$ and $2\sqrt{2}$.

My attempt is as follows:
There can be four cases, either $x$ lies in the first quadrant, second, third or fourth:-
First quadrant: $\cos x$ will decrease sharply and sinx will increase sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=0$
Second quadrant: $\cos x$ will increase in magnitude and sinx will decrease sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=\pi$
Third quadrant: $\cos x$ will decrease in magnitude and sinx will increase in magnitude but negative, so $y_{min}$ would tend to $-\infty$ near to $x=\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
Fourth quadrant: $\cos x$ will increase sharply and sinx will decrease in magnitude, so $y_{min}$ would tend to $-\infty$ near to $x=2\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
So in this way I have proved that $\dfrac{3+\cos x}{\sin x}$ cannot lie between $-2\sqrt{2}$ and $2\sqrt{2}$, but is their any smart solution so that we can calculate quickly.
 [/INST] If $\frac{3 + \cos x}{\sin x} = k, k \in \mathbb R$, then $3 + \cos x = k \sin x \Rightarrow 9 + 6 \cos x + \cos^2 x = k^2 - k^2 \cos^2 x$, thus $(k^2+1)\cos^2 x + 6 \cos x + (9 - k^2) = 0$. Let $u = k^2$. Now for no value of $\cos x$ to exist, the discriminant must be less than $0$:
$$6^2 - 4(u+1)(9-u) < 0 \Rightarrow 4(u+1)(9-u) > 36$$
$$\Rightarrow -u^2 + 8u + 9 > 9 \Rightarrow u(-u+8) > 0 \Rightarrow 0 < u < 8$$
where in the last step, a sketch of the quadratic shows that it is concave up, hence the direction of the inequality.
Thus $0 < k^2 < 8$. This implies there are no values of $x$ such that $-2 \sqrt{2} < \frac{3 + \cos x}{\sin x} < 2 \sqrt{2}$.
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