Zadanie

[ChiliGREEN]

\(\displaystyle{ \left tg^{2} x - 2tgx + 1 = 0}\)
\(\displaystyle{ \left cos^{2} x - 5cosx + 4 = 0}\)
\(\displaystyle{ \left sin^{2} x - 5sin + 6 = 0}\)
\(\displaystyle{ \left sin^{2} x - cos^{2} x -1 = 0}\)

[Hatcher]

a.\(\displaystyle{ tg^{2} x - 2tgx + 1 = 0}\)
Podstawiamy:\(\displaystyle{ t=tg{x}}\), pamiętając o tym, że:\(\displaystyle{ D_{tg}=R-[(\frac{\pi}{2}+k\pi), k C]}\)
\(\displaystyle{ t^2-2t+1=0 \iff (t-1)^2=0 \iff t=1}\)
Wracamy do podstawienia:
\(\displaystyle{ tg{x}=1}\), i rozwiązujemy ....


b.\(\displaystyle{ \left cos^{2} x - 5cosx + 4 = 0}\)
Niech \(\displaystyle{ t=\cos{x}}\),\(\displaystyle{ t }\)
\(\displaystyle{ t^2-5t+4=0 \iff t=1 t=4}\), jedno rozwiązanie odrzucamy
więc:\(\displaystyle{ cosx=1......}\)

c.\(\displaystyle{ \left sin^{2} x - 5sin + 6 = 0}\)
Niech \(\displaystyle{ t=sinx}\),\(\displaystyle{ t }\)
\(\displaystyle{ t^2-5t+6=0 \iff t=2 t=3}\) oba rozwiązania odrzucamy , a więc:\(\displaystyle{ Z_R=\o}\)


d.\(\displaystyle{ \left sin^{2} x - cos^{2} x -1 = 0}\)
\(\displaystyle{ sin^2{x}-1+sin^2{x}-1=0 \iff sin^2{x}=1 \iff sin{x}=1 sin{x}=-1 \iff ....}\)