If $x$ and $\alpha$ are real, then the inequation $\log_2x+\log_x2+2\cos\alpha \leq0$ has..? I have a question that is as follows:
If $x$ and $\alpha$ are real, then the inequation $\log_2x+\log_x2+2\cos\alpha \leq0$ has:
(a) no  solution
(b) has exactly two solutions
(c) is satisfied by any real $\alpha$ and any real $x \in (0,1)$
(d) is satisfied by any real $\alpha$ and any real $x \in (1,\infty )$
The answer is given as (c)
Here's what I have tried, and I cannot proceed further ..... :
The given inequation is equivalent to:
$\log_2x+\frac{1}{\log_2x}+2\cos\alpha \leq0$
Now, if I let $x \in (1,\infty)$ then clearly $\log_2x >0$ and, by AM-GM inequality, I get that
$\log_2x + \frac{1}{\log_2x} \geq 2$
I also know that $2\cos \alpha \in [-2,2]$
This is what I've managed to analyse and cannot proceed further, plz help.
 A: We have

$$\log_2x+\log_x2+2\cos\alpha \leq0, 1≠x>0.$$

If $\log_2(x)>0,$ then you get

$$\log_2x + \frac{1}{\log_2x} =u(x)≥2$$
$$u(x)+2\cos \alpha ≤0$$
$$-1≤\cos \alpha ≤- \frac {u(x)}{2}≤-1$$
$$\cos \alpha =-1, x=2$$

If $\log_2(x) <0 \Longleftrightarrow 0<x<1$, you get

$$\log_2x + \frac{1}{\log_2x} =-v(x), v(x)\geq 2$$
$$-v(x)+2\cos \alpha ≤0, v(x)≥2$$
$$\cos \alpha ≤\frac{v(x)}{2}, v(x)≥2$$

which is correct for any $\alpha \in\mathbb R$.
A: $\log_2 x < 0$ if $x < 1$ so AM.GM wouldn't apply.  But $\log_x 2 =\frac 1{\log_2 x} < 0$ so if $x < 1$ then we'd have $\log_x 2 + \log_2 x = -(|\log_x 2| + \frac 1{|\log_2x|}) \le -2$ and as $2\cos \theta \le 2$ we will have $\log_x 2 + \log_2 x + 2\cos \theta$ will have solutions for all $x \in (0,1)$.  And $\cos \theta$ may be anything so any $\theta$ will do.
So $c$ is true and $a$ and $b$ are false.
We're done if we assume exactly one of the options is true but to see why $d$ is false:
And if $x > 1$ then well, we have $\log_x 2 + \log_2 x \ge 2 \ge 2\cos \theta$ which can only be solved if $\log_x 2 + \log_2 x = 2 = 2\cos \theta$.
That is only true if $\log_x 2 = 1=\cos \theta$ or $x = 2$ and $\theta = 2k \pi$.
So solutions exist but not for every $x \in (1, \infty)$ and not for every $\theta$.
So $d$ is false.
We can go further.  There's no solutions for $x \le 0$ as $\log_2 x$ and $\log_x 2$ would not be define.
Every $x\in (0,1); \theta \in \mathbb R$ is a solutions.
If $x = 1$ there is not solution as $\log_1 2$ is not defined or meaningful.
The solutions in $x > 1$ are $x=2; \theta = 2k\pi$.
So solutions are $(x, \theta) \in (0,1)\times \mathbb R \cup \{(1,2k\pi)|k\in \mathbb N\}$
