cos2x + sin2x = 0

[server88]

Jak w temacie, zrobiłem tak:

\(\displaystyle{ \cos 2x + \sin 2x = 0}\)
\(\displaystyle{ 1 + \tan 2x = 0}\)
\(\displaystyle{ \tan 2x = -1}\)
\(\displaystyle{ \frac{2\tan x}{1-(\tan x)^{2}} = -1}\)
\(\displaystyle{ (\tan x)^{2} - 2\tan x - 1 = 0}\)
\(\displaystyle{ \tan x = t}\)
\(\displaystyle{ t^{2} - 2t - 1 = 0}\)
...
...
\(\displaystyle{ \tan x = 1 - \sqrt{2}}\)
\(\displaystyle{ \tan x = 1 + \sqrt{2}}\)

i nie wiem co dalej, czy w ogóle dobrze to jest?

[wb]

\(\displaystyle{ tg2x=-1 2x=- \frac{\pi}{4}+k\pi \ \ \ \ /:2 \ \\ x=- \frac{\pi}{8}+ \frac{k\pi}{2} \ \ \ ; k\in C}\)

[server88]

ok, dzięki, a tamto moje rozumowanie jest dobre?