Rozwiaz rownanie

[blantek22]

prosze o pomoc w rozwiazaniu bo nie moge dojsc jak sie je rozwiazuje:

1) tg�x-3=0
2) cos(2x-Π/6)-cos(x+Π/6=0
3) sin3x=sin(3x-Π/6)
4) cos�3x-1/2cos3x=0
5) ctg�x=3ctgx

z gory thx

[magdabp]

ad a)
\(\displaystyle{ tg^2x=3}\)
\(\displaystyle{ tgx=\sqrt{3}}\) lub \(\displaystyle{ tgx=-\sqrt{3}}\)
\(\displaystyle{ x=\frac{\pi}{3} + k\pi}\) lub \(\displaystyle{ x=-\frac{\pi}{3}+k\pi}\)

[ariadna]

5)
\(\displaystyle{ ctg^{3}x-3ctgx=0}\)
\(\displaystyle{ ctgx(ctg^{2}x-3)=0}\)
\(\displaystyle{ ctgx(ctgx-\sqrt{3})(ctgx+\sqrt{3})=0}\)
\(\displaystyle{ ctgx=0 ctgx=\sqrt{3} ctgx=-\sqrt{3}}\)

[pzielak]

2)

\(\displaystyle{ k C}\)

\(\displaystyle{ \cos (2x- \frac {\Pi}{6}) = \cos (x+ \frac {\Pi}{6})}\)
\(\displaystyle{ 2x- \frac {\Pi}{6} = x+ \frac {\Pi}{6} + 2k\Pi \ \ \ \ \ \ 2x- \frac {\Pi}{6} = -x- \frac {\Pi}{6} + 2k\Pi}\)
\(\displaystyle{ x = \frac {\Pi}{3} + 2k\Pi \ \ \ \ \ \ 3x = 2k\Pi}\)
\(\displaystyle{ x = \frac {\Pi}{3} + 2k\Pi \ \ \ \ \ \ x= \frac {2}{3}k\Pi}\)