3 | ||
sin2x= | ||
4 |
√3 | ||
sinx= | ||
2 |
π | 2π | |||
x= | +2kπ x= | +2kπ | ||
3 | 3 |
π | π + 6kπ | π(6k+1) | ||||
x = | + 2kπ = | = | < 50 π ... i rozwiąź | |||
3 | 3 | 3 |
2 | ||
x = | π + 2kπ = ... i analogicznie | |
3 |
3 | ||
sin2x = | ||
4 |
√3 | ||
sinx = − | . | |
2 |