| |x| | ||
a) sinx= | ||
| x |
| x−|x| | ||
b) cos | =1 | |
| 2 |
| x | ||
sinx = | ||
| x |
| π | ||
sinx = 1 ⇒ x = | + 2kπ ; k ∊ C ale pamietamy o tym, ze x ma być ≥ 0 zatem z tego | |
| 2 |
| π | ||
x = | + 2kπ ; k ∊ {0 ; 1 ; ...} | |
| 2 |
| −x | π | |||
sinx = | = −1 ⇒ x = − | + 2kπ ; k ∊ C ale pamiętamy o tym, że x < 0 więc | ||
| x | 2 |
| −π | ||
x = | − 2kπ ; k ∊ { 0 ; 1 ; ... } | |
| 2 |
| π | −π | |||
Ostatecznie x = | + 2kπ ; k ∊ {0 ; 1 ; ...} v x = | − 2kπ ; k ∊ { 0 ; 1 ; ... } | ||
| 2 | 2 |