?
cos2x−12sinx=12=0
wiem że to będzie tak;
1−sin2x−12sinx−12=0
−t2−t+12
Δ=√3
| 1−√3 | ||
x1= | ||
| −2 |
| 1+√3 | ||
x2= | ||
| −2 |
| 1 | ||
−t2 −t + | ? | |
| 2 |
| 1 | 1 | |||
−t2− | t+ | =0 /*(−2) i t∊<−1;1> | ||
| 2 | 2 |
| −1−3 | −1+3 | 1 | ||||
t1= | =−1 lubt2= | = | ||||
| 4 | 4 | 2 |
| 1 | ||
sinx=−1 lub sinx= | ||
| 2 |
| 3π | π | 5π | ||||
x= | +2kπ lub x= | +2kπ lub x= | +2kπ | |||
| 2 | 6 | 6 |