Solve: $\sec(2x) \ge\sec(x) , x\in [0,\pi]$\ {$\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$} Here is the following question:
Solve: $\sec(2x) \ge\sec(x) , x\in [0,\pi]$\
{$\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$}
Note: This is part b of a question where in part a, I was asked to solve: $\cos(2x)=\cos(x), x\in[0,2\pi]$
For part a, I got my answers as $x = 0,\frac{2\pi}{3},\frac{4\pi}{3},2\pi$
Part a aided me as I was able to get part b to the following form:

*

*$\cos(x)>=\cos(2x)$
However, here is where I am confused. Are the answers just $x>=0, x>=\frac{2\pi}{3}$ by copying the signs?
 A: Recall that $\sec(x) = \frac{1}{\cos x}$ and $\cos(2x) = 2 \cos^2(x) - 1$.  So, in terms of $\cos(x)$, you have:
$$\frac{1}{2 \cos^2(x) - 1} \ge \frac{1}{\cos(x)}$$
For convenience, let $c = \cos(x)$.
$$\frac{1}{2 c^2 - 1} \ge \frac{1}{c}$$
Now, cross-multiply.  Note that because $-1 \le c \le 1$, $2c^2 - 1$ is always positive.  But $c$ can be either positive or negative.  (It can't be zero, because then the division on the RHS is undefined.)  If it's positive, then:
$$c \ge 2c^2 - 1$$
$$0 \ge 2c^2 - c - 1$$
$$0 \ge (2c + 1)(c - 1)$$
Since $0 < c \le 1$, $2c + 1$ is positive, and $c - 1$ is nonpositive.  Thus, the inequality is true for all $0 < c \le 1$.
OTOH, if $c < 0$, then we need to flip the inequality sign in all the steps above, so $$0 \le (2c + 1)(c - 1)$$
Since $-1 \le c < 0$, $2c + 1$ is positive, and $c - 1$ is strictly negative, so $(2c + 1)(c - 1)$ is negative, and the inequality is not satisfied.
Therefore, the original inequality is true if $0 < \cos(x) \le 1$.  Since the domain of $x$ is restricted to $[0, \pi] ∖ \{ \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4} \}$, this means that $x \in (0, \frac{\pi}{2}) ∖ {\frac{\pi}{4}}$.
