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 |
π | ||
sinx = sin | ||
5 |