# What are the all the solutions between 0 and 2π for sin2x-1=0?

Mar 7, 2016

$x = \frac{\pi}{4}$ or $x = \frac{5 \pi}{4}$

#### Explanation:

$\sin \left(2 x\right) - 1 = 0$

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

$\sin \left(\theta\right) = 1$ if and only if $\theta = \frac{\pi}{2} + 2 n \pi$ for $n \in \mathbb{Z}$

$\implies 2 x = \frac{\pi}{2} + 2 n \pi$

$\implies x = \frac{\pi}{4} + n \pi$

Restricted to $\left[0 , 2 \pi\right)$ we have $n = 0$ or $n = 1$, giving us

$x = \frac{\pi}{4}$ or $x = \frac{5 \pi}{4}$

Mar 7, 2016

$S = \left\{\frac{\pi}{4} , 5 \frac{\pi}{4}\right\}$

#### Explanation:

First, isolate the sine

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

Now, take a look at your unit circle Now, the sine correspond to the $y$ axis, so we can see that the only point between $0$ and $2 \pi$ where the sine is $1$ is $\frac{\pi}{2}$ radians, so we have:

$2 x = \frac{\pi}{2}$

We want to solve for x, so

$x = \frac{\pi}{4}$

However, remember that the period of the normal sine wave is $2 \pi$, but since we're working with $\sin \left(2 x\right)$, the period has changed; basically what we know is that there is a constant $k$ that will act as the period, so:

$2 \left(\frac{\pi}{4} + k\right) = \frac{\pi}{2} + 2 \pi$
$\frac{\pi}{2} + 2 k = \frac{\pi}{2} + 2 \pi$
$2 k = 2 \pi$
$k = \pi$

And since $\frac{\pi}{4} + \pi$ or $5 \frac{\pi}{4}$ is between $0$ and $2 \pi$, that enters our set of solutions.

$S = \left\{\frac{\pi}{4} , 5 \frac{\pi}{4}\right\}$