| π | ||
Rozwiąż rówanie 2cos(3x− | )=1 dla x ∊ <0,2π> | |
| 4 |
| π | 1 | |||
2cos2(3x− | )= | o krawdracie zapomnialem | ||
| 4 | 2 |
| π | ||
cos2(3x− | )=1 o tak  na pewno | |
| 4 |
| π | ||
2cos2(3x− | )=1 omg. jednak ta wersja na bank... | |
| 4 |
| π | 1 | |||
cos2(3x− | )= | /√ | ||
| 4 | 2 |
| π | √2 | π | √2 | |||||
cos(3x− | )= | v (3x− | )=− | |||||
| 4 | 2 | 4 | 2 |