Mam nadzieję, że ktoś pomoże
a) cos(2x−π6)−cos(x+π6)=0
b) tg2x=tg(3x−π6)
c) sin3x=sin(x+π4)
d) cos23x−12cos3x=0
e) sin2(12x)+1=2sin(12x)
f) ctg3=3ctgx
| π | π | |||
cos(2x − | = cos(x + | ) | ||
| 6 | 6 |
| π | π | |||
2x − | = x + | + 2kπ | ||
| 6 | 6 |
| π | ||
x = | + 2kπ | |
| 3 |
| π | ||
tg2x = tg(3x − | ) | |
| 6 |
| π | ||
2x = 3x − | + kπ | |
| 6 |
| π | ||
x = | − kπ | |
| 6 |
| π | ||
3x = x + | + 2kπ | |
| 4 |
| π | ||
2x = | + 2kπ | |
| 4 |
| π | ||
x = | + kπ | |
| 8 |
| 1 | ||
cos3x(cos3x − | ) = 0 | |
| 2 |
| 1 | ||
cos3x = 0 v cos3x = | ||
| 2 |
| π | π | π | π | |||||
3x = | + 2kπ v x = − | + 2kπ v 3x = | + 2kπ v x = − | + 2kπ | ||||
| 2 | 2 | 3 | 3 |
| 1 | ||
sin( | x) = t | |
| 2 |
| 1 | ||
sin( | x) = 1 | |
| 2 |
| 1 | π | ||
x = | + 2kπ | ||
| 2 | 2 |
| π | π | π | π | |||||
2x − | = x + | + 2kπ v 2x − | = −x − | + 2kπ | ||||
| 6 | 6 | 6 | 6 |
| π | 2 | |||
x = | + 2kπ v 3x = 2kπ => x = | kπ | ||
| 3 | 3 |
| π | 2 | |||
x = | + 2kπ v x = | kπ | ||
| 3 | 3 |
Prosiłbym także o wyjaśnienie e), bo kompletnie nie rozumiem tego zapisu
| π | π | |||
3x = x + | + 2kπ v 3x = π − (x + | ) + 2kπ | ||
| 4 | 4 |
| π | 3 | |||
2x = | + 2kπ v 4x = | π + 2kπ | ||
| 4 | 4 |
| π | 3 | 1 | ||||
x = | + kπ v x = | π + | kπ | |||
| 8 | 16 | 2 |
| π | π | |||
sprzeczność v x = | + kπ v x = − | + kπ | ||
| 6 | 6 |
| 1 | ||
cosx = | wtedy rozpisuje się to na: | |
| 2 |
| π | π | |||
x = | + 2kπ v x = − | + 2kπ | ||
| 3 | 3 |
| 1 | ||
sinx = | ||
| 2 |
| π | π | |||
x = | + 2kπ v x = π − | + 2kπ | ||
| 6 | 6 |
Wielkie dzięki za ekspresową pomoc!