| π | ||
cosx−cos(x+ | )= 0 | |
| 3 |
| 2x + π/3 | −π/3 | |||
−2sin | sin | =0 | ||
| 2 | 2 |
| π | −π | |||
−2sin(x + | ) sin | =0 | ||
| 6 | 6 |
| π | −π | |||
sin(x + | )= 0 v sin | =0 | ||
| 6 | 6 |
| π | ||
x+ | =kπ | |
| 6 |
| π | ||
x=− | +kπ taki wynik jest w książce | |
| 6 |
| −π | ||
sin | =0 | |
| 6 |
| π | 1 | |||
sin(− | ) = − | . | ||
| 6 | 2 |
| − π | ||
i wtedy mam | = 0+2kπ ale co z tym zrobić dalej | |
| 6 |
| −π | 1 | −π | ||||
Skoro sin | = − | to warunek sin | =0 jest jakby trochę nieprawdziwy ![]() | |||
| 6 | 2 | 6 |