# Find all solutions on the interval [0, 2pi)? 2sin(2x)-1=0

Apr 3, 2018

$x = \frac{\pi}{12} , x = \frac{5 \pi}{12} , x = \frac{13 \pi}{12} , x = \frac{17 \pi}{12}$

#### Explanation:

We start by adding 1 to both sides:

$2 \sin \left(2 x\right) - \cancel{1 + 1} = 0 + 1$

$2 \sin \left(2 x\right) = 1$

Next we divide both sides by $2$:

$\frac{\cancel{2} \sin \left(2 x\right)}{\cancel{2}} = \frac{1}{2}$

$\sin \left(2 x\right) = \frac{1}{2}$

We know that $\sin \left(\frac{\pi}{6} + 2 \pi k\right) = \frac{1}{2}$, which gives:

$2 x = \frac{\pi}{6} + 2 \pi k$

$\frac{\cancel{2} x}{\cancel{2}} = \left(\frac{\frac{\pi}{6} + 2 \pi k}{2}\right)$

$x = \frac{\pi}{12} + \pi k$

This is one of the solutions. We can find the other by using the identity $\sin \left(\theta\right) = \sin \left(\pi - \theta\right)$. This means that $\sin \left(\pi - \frac{\pi}{6} + 2 \pi k\right)$ also equals $\frac{1}{2}$, which gives the solution:

$2 x = \pi - \frac{\pi}{6} + 2 \pi k$

$2 x = \frac{5 \pi}{6} + 2 \pi k$

$\frac{\cancel{2} x}{\cancel{2}} = \frac{5 \pi}{2 \cdot 6} + \frac{2 \pi k}{2}$

$x = \frac{5 \pi}{12} + \pi k$

Now we need to find which of these solutions are inside the desired interval. We can write them out, seeing which ones are smaller than $2 \pi$:

$\frac{\pi}{12} + \pi k \to \frac{\pi}{12} , \frac{13 \pi}{12} , \frac{25 \pi}{12.} . .$

$\frac{5 \pi}{12} + \pi k \to \frac{5 \pi}{12} , \frac{17 \pi}{12} , \frac{29 \pi}{12}$

Everything larger than $\frac{24 \pi}{12} = 2 \pi$, is not in the interval we want, so we can ignore those, leaving us with the following solutions:

$x = \frac{\pi}{12} , x = \frac{5 \pi}{12} , x = \frac{13 \pi}{12} , x = \frac{17 \pi}{12}$