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Letty: Rozwiąż równianie:
cos22x + 4cos2x = 2
14 maj 17:29
Basia:
cos2x = cos
2x − sin
2x = 2cos
2x − 1
(2cos
2x−1)
2 + 4cos
2x − 2 = 0
4cos
4x − 4cos
2x + 1 + 4cos
2x − 2 =0
4cos
4x − 1 = 0
(2cos
2x−1)(2cos
2x+1) = 0
| | 1 | | 1 | |
cos2x = |
| lub cos2x= − |
| (niemożliwe) |
| | 2 | | 2 | |
cos
2x −
12 = 0
| | √1 | | 1 | | √2 | |
√12 = |
| = |
| = |
| |
| | √2 | | √2 | | 2 | |
| | √2 | | √2 | |
(cosx − |
| )(cosx+ |
| ) = 0 |
| | 2 | | 2 | |
dokończ
14 maj 18:23