jak rozwiązać takie równanie
4sinxcosx−√3=0
Czyli:
4sinxcosx=2*2sinxcosx=2sin2x
2sin2x=√3
| √3 | ||
sin2x= | ||
| 2 |
| √3 | ||
wiem że sin2x= | ||
| 2 |
| π | ||
i ze x= | +kπ | |
| 6 |
| 2 | ||
to drugi będzie x= | π+2kπ | |
| 3 |
| π | ||
2x = | + 2kπ | |
| 3 |
| π | ||
2x = π − | + 2kπ | |
| 3 |
| 2π | ||
2x = | + 2kπ | |
| 3 |
| π | ||
x = | + kπ | |
| 3 |
| π | π | |||
x = { | + kπ, | + kπ}, k ∊ C | ||
| 6 | 3 |
| 1 | ||
sinπx = | ||
| 2 |
| π | ||
πx = | + 2kπ | |
| 6 |
| π | ||
πx = π − | + 2kπ | |
| 6 |
| 1 | 5 | |||
x = { | + 2k, | + 2k}.  | ||
| 6 | 6 |