6 mar 18:47
Basia:
3
sin2x = 2 + 3
1−sin2x
| | 3 | |
3sin2x = 2 + |
| /*3sin2x |
| | 3sin2x | |
(3
sin2x)
2 = 2*3
sin2x + 3
t = 3
sin2x
0 ≤ sin
2x ≤ 1
3
0 ≤ t ≤ 3
1
0 ≤ t ≤ 3
i masz
t
2 = 2t+3
dokończ
6 mar 18:55
Mila:
3
sin2x=2+3
cos2x
3
sin2x−3
1−sin2x−2=0
3
sin2x−3*3
−sin2x−2=0
3
sin2x=t, t>0
t
2−2t−3=0
t=−1∉D lub t=3
3
sin2x=3⇔sin
2x=1
sinx=1 lub sinx=−1
6 mar 19:00
