| 1 | ||
Jak rozwiązć tę nierówność (1−cosx)(1+cos2x)(1−cos3x)< | ||
| 2 |
| 1 | ||
(1−cosx)*2cos2x*(1−cos3x)< | ⇔ | |
| 2 |
| 1 | ||
(1−cosx)*cos2x*(1−cos3x)< | ⇔ | |
| 4 |
| 1 | ||
(1−cosx)*cos2x*(1−4cos3x+3cosx)− | <0 | |
| 4 |
| 1 | ||
(1−t)*t2*(1−4t3+3t)− | <0⇔ | |
| 4 |
| 1 | ||
t2*(t−1)*(4t3−3t−1)− | <0⇔ | |
| 4 |
| 1 | ||
t2*(t−1)*(t−1)*(2t+1)2− | <0 | |
| 4 |
| 1 | ||
[t*(t−1)*(2t+1)]2−( | )2<0 | |
| 2 |
| 1 | 1 | |||
[t*(t−1)*(2t+1)− | ]*[t*(t−1)*(2t+1)+ | ]<0 | ||
| 2 | 2 |