1 ) Solve in R this equation: ( let:tan x= t)
2 )Deduct the values of :
$\displaystyle 2\sin{x}\cos{x} = \sin{(2x)}$.
So if $\displaystyle \cos{(2x)} + 2\sin{x}\cos{x} = 0$
$\displaystyle \cos{(2x)} + \sin{(2x)} = 0$
$\displaystyle \cos{(2x)} = -\sin{(2x)}$
$\displaystyle 1 = -\frac{\sin{(2x)}}{\cos{(2x)}}$
$\displaystyle -\tan{(2x)} = 1$
$\displaystyle 2x = \left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n$, where $\displaystyle n \in \mathbf{Z}$
$\displaystyle x = \frac{1}{2}\left[\left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n\right]$
$\displaystyle x = \left\{\frac{\pi}{8}, \frac{5\pi}{8}\right\} + \pi n$, where $\displaystyle n \in \mathbf{Z}$.
I've just realised I made a mistake...
I got to the step
$\displaystyle -\tan{(2x)} = 1$
$\displaystyle \tan{(2x)} = -1$
$\displaystyle 2x = \left\{\pi - \frac{\pi}{4}, 2\pi - \frac{\pi}{4}\right\} + 2\pi n, n \in \mathbf{Z}$
$\displaystyle 2x = \left\{\frac{3\pi}{4}, \frac{7\pi}{4}\right\} + 2\pi n$
$\displaystyle x = \frac{1}{2}\left[\left\{\frac{3\pi}{4}, \frac{7\pi}{4}\right\} + 2\pi n\right]$
$\displaystyle x = \left\{\frac{3\pi}{8}, \frac{7\pi}{8}\right\} + \pi n, n \in \mathbf{Z}$.
There was nothing wrong with my solution process, I just forgot about the negative sign.
You don't need to use the other method...