3Silnia&6: | | sinxcosx + 1 | | sinx | |
( |
| )2 = 1/3 ⇒ ( |
| )2 = 3 |
| | sinx | | sinxcosx +1 | |
sin
2x= 3(sin
2xcos
2x + 2sinxcosx + 1), sin
2x = t , 0 < t ≤ 1
t = 3( t(1−t) + 2
√t(1−t) + 1) ⇔ 3t
2 − 2t − 3 = 2
√t(1−t) / ()
2
(3t
2 − 2t − 3)
2 = 4t(1−t) ⇔ 9t
4 − 12t
3 − 10t
2 + 8t + 9 = 0 − wychodzi brak rozwiazan w R