<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] Squaring an equation may introduce unwanted roots, so it is generally avoided. Rewrite the equation, instead, as
$$p(x) = \frac{\sin x +\cos x}{\frac12\sin 2x} =\frac{2\sqrt2 \cos(x-\frac\pi4)}{2\cos^2(x-\frac \pi4)-1}$$
So, $p(x)$ is a function of $\cos(x-\frac\pi4)$ and symmetric around $\frac\pi4$ and $\frac{5\pi}4$, which are also the two local extrema over $(0,2\pi)$, i.e.
$$p_{min}(\frac\pi4) = 2\sqrt2,\>\>\>\>\>p_{max}(\frac{5\pi}4) = -2\sqrt2$$
Thus, for any $p$ satisfying $p^2>(\pm 2\sqrt2)^2=8$, it meets the curve $p=\sec x+ \csc x$ four times, i.e. four distinctive roots in the shaded area of the graph below,

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