Find the values of $sin$, $cos$, and $ an$ for angles that satisfy the equation $2sin(x)cos(x) = 1$

First, recognize that $2\sin(x)\cos(x) = \sin(2x)$. Thus, the equation becomes:

$\sin(2x) = 1$

The solution for $\sin(2x) = 1$ occurs at:

$2x = \frac{\pi}{2} + 2k\pi$, where $k$ is any integer.

Thus:

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

For $k = 0$:

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

Then:

$\sin(x) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

$\cos(x) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

$\tan(x) = \tan\left(\frac{\pi}{4}\right) = 1$

For $k = 1$:

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

Then:

$\sin(x) = \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

$\cos(x) = \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

$\tan(x) = \tan\left(\frac{5\pi}{4}\right) = 1$

First, recognize that $2sin(x)cos(x) = sin(2x)$. Thus, the equation becomes:

$sin(2x) = 1$

The solution for $sin(2x) = 1$ occurs at:

$2x = frac{pi}{2} + 2kpi$, where $k$ is any integer.

Thus:

$x = frac{pi}{4} + kpi$

For $k = 0$:

$x = frac{pi}{4}$

Then:

$sin(x) = frac{sqrt{2}}{2}$

$cos(x) = frac{sqrt{2}}{2}$

$ an(x) = 1$

For $k = 1$:

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

Then:

$sin(x) = -frac{sqrt{2}}{2}$

$cos(x) = -frac{sqrt{2}}{2}$

$ an(x) = 1$

Using the identity $2sin(x)cos(x) = sin(2x)$, we have:

$sin(2x) = 1$

Thus, $2x = frac{pi}{2} + 2kpi$, so:

$x = frac{pi}{4} + kpi$

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