# How to solve this inequality |sin2x|>sqrt3/2  ?

## $| \sin 2 x | > \frac{\sqrt{3}}{2}$

Jan 31, 2016

$\ldots \cup \left(- \frac{5 \pi}{6} , - \frac{2 \pi}{3}\right) \cup \left(- \frac{\pi}{3} , - \frac{\pi}{6}\right) \cup \left(\frac{\pi}{6} , \frac{\pi}{3}\right) \cup \left(\frac{2 \pi}{3} , \frac{5 \pi}{6}\right) \cup \left(\frac{7 \pi}{6} , \frac{4 \pi}{3}\right) \cup \left(\frac{5 \pi}{3} , \frac{11 \pi}{6}\right) \cup \ldots$

#### Explanation:

We can say more generally that we want to find the values of $x$ which satisfy

$\frac{\sqrt{3}}{2} < \sin 2 x$ or $\sin 2 x < \frac{\sqrt{3}}{2}$.

The period of $\sin \left(2 x\right)$ is $\pi$, i.e. it repeats itself after an interval of $\pi$. Mathematically, we write $\sin \left(2 x\right) = \sin \left(2 x + \pi\right)$ for any $x \in \mathbb{R}$. A graph of $y = \sin \left(2 x\right)$ is appended below.

graph{sin(2x) [-10, 10, -5, 5]}

Therefore, we will limit $x \in \left[0 , \pi\right)$ first.

We first attempt to solve the first inequality.

$\frac{\sqrt{3}}{2} < \sin 2 x$

Since $\sin$ is positive in the first and second quadrant, and we know that the basic angle is ${\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)$, we get

${\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right) < 2 x < \pi - {\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)$

$\frac{\pi}{3} < 2 x < \pi - \frac{\pi}{3}$

$\frac{\pi}{3} < 2 x < \frac{2 \pi}{3}$

$\frac{\pi}{6} < x < \frac{\pi}{3}$

For the second inequality,

$\sin 2 x < \frac{\sqrt{3}}{2}$

Since $\sin$ is negative in the third and forth quadrant, we get

$\pi + {\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right) < 2 x < 2 \pi - {\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)$

$\pi + \frac{\pi}{3} < 2 x < 2 \pi - \frac{\pi}{3}$

$\frac{4 \pi}{3} < 2 x < \frac{5 \pi}{3}$

$\frac{2 \pi}{3} < x < \frac{5 \pi}{6}$

After that we combine the two solution sets together (union) and we get for $x \in \left[0 , \pi\right)$, the solution set is

$\left(\frac{\pi}{6} , \frac{\pi}{3}\right) \cup \left(\frac{2 \pi}{3} , \frac{5 \pi}{6}\right)$.

For $x \in \mathbb{R}$, you just have to add integer multiples of $\pi$. So the final answer looks like

$\ldots \cup \left(- \frac{5 \pi}{6} , - \frac{2 \pi}{3}\right) \cup \left(- \frac{\pi}{3} , - \frac{\pi}{6}\right) \cup \left(\frac{\pi}{6} , \frac{\pi}{3}\right) \cup \left(\frac{2 \pi}{3} , \frac{5 \pi}{6}\right) \cup \left(\frac{7 \pi}{6} , \frac{4 \pi}{3}\right) \cup \left(\frac{5 \pi}{3} , \frac{11 \pi}{6}\right) \cup \ldots$

Jan 31, 2016

(pi/6, pi/3)
(2pi/3, (5pi)/6)

#### Explanation:

Separate the inequality in 2 parts:
a. $\sin 2 x > \frac{\sqrt{3}}{2}$ and
b. $- \sin 2 x > \frac{\sqrt{3}}{2}$ --> $\sin 2 x < - \frac{\sqrt{3}}{2}$
a. Solve $\sin 2 x > \frac{\sqrt{3}}{2}$ by the trig unit circle.
The inequality is true when the arc 2x varies between $\frac{\pi}{3}$ and $\frac{2 \pi}{3}$ on the unit circle
$\frac{\pi}{3} < 2 x < \frac{2 \pi}{3}$
$\frac{\pi}{6} < x < \frac{2 \pi}{6} = \frac{\pi}{3}$
b. Solve $\sin 2 x < - \frac{\sqrt{3}}{2}$.
The inequality is true when the arc 2x varies between $\frac{4 \pi}{3}$ and $\frac{5 \pi}{3} :$ on the unit circle. In this interval, $\sin 2 x < - \frac{\sqrt{3}}{2}$
$\frac{4 \pi}{3} < 2 x < \frac{5 \pi}{3}$
$\frac{2 \pi}{3} < x < \frac{5 \pi}{6}$