| 1 | 1 | 1 | ||||
a) założenie: x + | π ≠ | π + kπ ⇒ x ≠ | π + kπ. | |||
| 3 | 2 | 6 |
| 1 | ||||||||||||
4* | *cos2(x+ | π) = √3 / :2 | |||||||||||
| 3 |
| 1 | 1 | √3 | ||||
2sin(x+ | π) cos(x+ | π) = | ||||
| 3 | 3 | 2 |
| 2 | 1 | |||
sin(2x + | π) = sin( | π) | ||
| 3 | 3 |
| 2 | 1 | 2 | 1 | |||||
2x + | π = | π + k*2π lub 2x + | π = π − | π + k*2π | ||||
| 3 | 3 | 3 | 3 |
| 1 | ||
2x = − | π + k*2π lub 2x = k*2π | |
| 3 |
| 1 | ||
x = − | π + k*π lub x = k*π | |
| 6 |
| α + β | α − β | |||
b) zastosuj wzór: cosα + cosβ = 2cos | cos | |||
| 2 | 2 |