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.
 A: Use the half-angle expressions 
$\cos x = \frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}$ and 
$\sin x = \frac{2\tan\frac x2}{1+\tan^2\frac x2}$ to express 
$$I=\frac{3+\cos x}{\sin x}=
\frac{2}{\tan\frac x2} +\tan \frac x2$$
Note 
$$I^2=\left(\frac{2}{\tan \frac x2} +\tan \frac x2\right)^2
=\left(\frac{2}{\tan \frac x2} -\tan \frac x2\right)^2+8 \ge 8$$
Thus, $I^2$ can not have values within $[0,8)$, which means that $I=\frac{3+\cos x}{\sin x}$ can not have values within $(-2\sqrt2, \>2\sqrt2)$.
A: Let the value of this expression at $x=p$ be $q$. Then we have
$${3+\cos p\over\sin p}= q$$
$$\implies\cos p-q\sin p = -3$$
Dividing both LHS and RHS by $\sqrt{q^2+1}$ we have 
$$\cos p\cdot {1\over\sqrt{q^2+1}} - \sin p\cdot{q\over\sqrt{q^2+1}} = {-3\over\sqrt{q^2+1}}$$
Let $1\over\sqrt{q^2+1}$ be $\cos r$. So, we have $$\cos p\cos r - \sin p\sin r = {-3\over\sqrt{q^2+1}}$$
Or $$\cos{(p+r)} = \frac{-3}{\sqrt{q^2+1}}$$
For this to be a valid expression $\sqrt{q^2+1}$ must be greater than $3$ since $\cos x \in [-1,1]$. So, we have 
$$q^2+1 \geq 9$$
$$\implies |q| \geq \sqrt 8$$
$$\implies q \in \left(-\infty,-2\sqrt 2\right]\cup\left[2\sqrt 2, \infty\right)$$
$$\implies\boxed{ {3+\cos x\over\sin x}\notin\left(-2\sqrt 2, 2\sqrt 2\right)}$$
A: Basically you want to prove:
$$\left|\dfrac{3+\cos x}{\sin x}\right|\ge 2\sqrt{2} \iff (3+\cos x)^2\ge 8\sin ^2x \iff (3\cos x+1)^2\ge 0 \ \ \checkmark$$
Note: Equality occurs for $\cos x=-\frac13 \Rightarrow \sin x=\pm \frac{2\sqrt2}{3}$.
A: 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 $-2 \sqrt{2} < \frac{3 + \cos x}{\sin x} < 2 \sqrt{2}$.
