| π | π | |||
a) 3sin(x− | )+cos (x+ | )=1 | ||
| 4 | 4 |
| π | ||
b) sinx=sin | ||
| 5 |
| π | π | π | π | |||||
3(sinxcos | −cosxsin | )+cosxcos | −sinxsin | = 1 /*√2 | ||||
| 4 | 4 | 4 | 4 |
| √2 | ||
2sinx−2cosx = √2 |* | ||
| 4 |
| π | π | 1 | ||||
sinx*cos | −cosx*sin | = | ||||
| 4 | 4 | 2 |
| π | 1 | |||
sin(x− | ) = | ....dokończ.... | ||
| 4 | 2 |
| π | ||
sinx = | ||
| 5 |
| π | π | |||
x = | +2kπ lub x = (π− | )+2kπ | ||
| 5 | 5 |
Poprawka
| π | ||
sinx = sin | ||
| 5 |
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