<s>[INST] Trigonometric Inequality $\sin (2x) \gt \sqrt 2 \sin (x)$ I wish to solve this inequality:
$\sin (2x) \gt \sqrt 2 \sin (x)$
My approach:
I tried to isolate the $x$ on the left side by using the sine sum formula:
$2\sin(x)\cos(x) \gt \sqrt2\sin(x)$
then I divided by $\sin(x) \over 2$ both sides:
$\cos(x) \gt {\sqrt2 \over2}$
$x \lt \cos^{-1}({\sqrt2 \over2})$
$x < {\pi \over4}$
From that I can conclude that $x < {7\pi \over4}$, but I know the answer is still incomplete as it should be 
$0 \lt x \lt {\pi \over 4}$, $\pi \lt x \lt {7 \over 4}\pi$
As I was able to see on Desmos graph plotter.
Does my approach gives the tools to reach this answer or have I commited a mistake?
 [/INST] You should not have divided by $\sin x$.  That causes you to lose information you need to solve the problem.
\begin{align*}
\sin(2x) & > \sqrt{2}\sin x\\
2\sin x\cos x & > \sqrt{2}\sin x\\
2\sin x\cos x - \sqrt{2}\sin x & > 0\\
\sqrt{2}\sin x(\sqrt{2}\cos x - 1) & > 0
\end{align*}
The inequality is satisfied if $\sin x > 0$ and $\sqrt{2}\cos x - 1 > 0$ or if $\sin x < 0$ and $\sqrt{2}\cos x - 1 < 0$.
Let's focus on solving the problem in the interval $[0, 2\pi)$ for the moment.
The inequality $\sin x > 0$ is satisfied in the interval $[0, 2\pi)$ if $x \in (0, \pi)$.  The inequality $$\sqrt{2}\cos x - 1 > 0 \iff \cos x > \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ is satisfied in the interval $[0, 2\pi)$ if $x \in [0, \frac{\pi}{4}) \cup (\frac{7\pi}{4}, 2\pi)$. Hence, $\sin x > 0$ and $\sqrt{2}\cos x - 1 > 0$ if 
$$x \in (0, \pi) \cap \left\{\left[0, \frac{\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right)\right\} = \left(0, \frac{\pi}{4}\right)$$ 
The inequality $\sin x < 0$ is satisfied in the interval $[0, 2\pi)$ if $x \in (\pi, 2\pi)$.  The inequality $$\sqrt{2}\cos x - 1 < 0 \iff \cos x < \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ is satisfied in the interval $[0, 2\pi)$ if $x \in (\frac{\pi}{4}, \frac{7\pi}{4})$.  Hence, $\sin x < 0$ and $\sqrt{2}\cos x - 1 < 0$ if
$$x \in (\pi, 2\pi) \cap \left(\frac{\pi}{4}, \frac{7\pi}{4}\right) = \left(\pi, \frac{7\pi}{4}\right)$$
Thus, $\sin(2x) > \sqrt{2}\sin x$ in the interval $[0, 2\pi)$ if 
$$x \in  \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{7\pi}{4}\right)$$
Since the sine function has period $2\pi$, the general solution is 
$$x \in \bigcup_{k \in \mathbb{Z}} \left\{\left(2k\pi, \frac{\pi}{4} + 2k\pi\right) \cup \left(\pi + 2k\pi, \frac{7\pi}{4} + 2k\pi\right)\right\}$$ 
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