Trygonometria
Pati18773: Dobrze to przeksztalcilam ?
√3cos(x)+sin(x)=√2
√3cos(x)=√2−sin(x) /2
3cos2(x)=2−2√2sin(x)+sin2(x)
3(1−sin2(x))=2−2√2sin(x)+sin2(x)
3−3sin2(x)=2−2√2sin(x)+sin2(x)
−4sin2(x)+2√2sin(x)+1=0
12 gru 22:26
Eta:
Dzielimy obustronnie przez 2
| √3 | | 1 | | √2 | |
| *cosx+ |
| *sinx= |
| |
| 2 | | 2 | | 2 | |
............................
dokończ
12 gru 22:33
Mila:
√3cos(x)+sin(x)=√2 /:2
| √3 | | 1 | | √2 | |
| *cosx+ |
| *sinx= |
| |
| 2 | | 2 | | 2 | |
| | π | | π | | √2 | |
sin |
| *cosx+sinx*cos |
| = |
| ⇔ |
| | 3 | | 3 | | 2 | |
| | π | | π | | π | | 3π | |
x+ |
| = |
| +2kπ lub x+ |
| = |
| +2kπ |
| | 3 | | 4 | | 3 | | 4 | |
| | π | | 5π | |
x=− |
| +2kπ lub x= |
| +2kπ |
| | 12 | | 12 | |
12 gru 22:34
Eta:
12 gru 22:35
Pati18773: | | π | | π | | π | |
Dobra tylko ze przyrownalam sobie |
| −x= |
| −2kπ i wyszlo mi x=− |
| −2kπ |
| | 6 | | 4 | | 12 | |
12 gru 22:36
Pati18773: Tam zamiast − powinien byc +
12 gru 22:37
Pati18773: Dobrze rozumiem teraz dziekuje !
12 gru 22:37