<s>[INST] If $\sec x + \csc x =p$ has four distinct solutions between $(0,2\pi)$, then which if the following is incorrect? 
a) $p^2-8>0$
b) $p=\sqrt 2$
c) $p=-\sqrt 2$
d) $p=0$

My attempt
$$\frac{\sec x +\csc x}{2} \ge \sqrt {\sec x \csc x}$$
$$\frac{\sin x +\cos x}{\sin x \cos x }\ge 2\sqrt {\frac{1}{\sin x \cos x}}$$
$$\sin x +\cos x \ge 2\sqrt {\sin x \cos x}$$
$$(\sqrt {\sin x}-\sqrt {\cos x})^2\ge 0$$
$$\sin x \ge \cos x$$
I realise that some of that squaring might have removed or added some roots, but I don’t know what else to do
From this result, the interval for $x$ is $[\frac{\pi}{4}, \pi]\cup [\frac{5\pi}{4}, \frac{3\pi}{2}]$
Don’t know what do next. Can I get some insight?
Another attempt
$$\sin x +\cos x =p \sin x \cos x$$
$$1+2\sin x\cos x =\frac 14 p^2 4\sin^2x \cos ^2x$$
$$1+\sin 2x =\frac 14 p\sin^2 2x$$
$$p\sin^22x-4\sin 2x -4=0$$
 [/INST] The derivative of $p$ is equal to $\dfrac{\sin(x)}{\cos^2(x^2)}-\dfrac{\cos(x)}{\sin^2(x)}$ so$p$ has a minimum at $x=\dfrac{\pi}{4}$ and a maximum at $x=\dfrac{5\pi}{4}$. Consequently $p$ has four values in $[0,2\pi]$ for $\sec x + \csc x \gt2\sqrt2$ and for
$\sec x + \csc x \lt-2\sqrt2$.
Thus only $p^2-8\gt0$ is correct.
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