# 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?

• Are you looking for the general solution or the solution within a specified interval? – N. F. Taussig May 23 '18 at 0:14
• "then I divided by $\sin(x) \over 2$ both sides:" Ouch! What if the sine term is negative? – imranfat May 23 '18 at 0:18
• Dividing is sometimes problematic in inequalities. Instead, you may want to solve $\sqrt 2 \sin x ( \sqrt 2 \cos x - 1)>0$ – ThePortakal May 23 '18 at 0:27

The inequation can be written as

$$\sin (x)\Bigl(\cos (x)-\cos (\frac {\pi}{4})\Bigr)>0$$

which gives

$$2k\pi <x <\frac {\pi}{4}+2k\pi$$ or

$$(2k-1)\pi <x <2k\pi-\frac {\pi}{4}$$

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 $$x \in \bigcup_{k \in \mathbb{Z}} \left\{\left(2k\pi, \frac{\pi}{4} + 2k\pi\right) \cup \left(\pi + 2k\pi, \frac{7\pi}{4} + 2k\pi\right)\right\}$$