π | ||
a) f(x) = 2cos(x + | ) | |
3 |
2 π | ||
b) x = 2k π lub x = − | + 2k π | |
3 |
1 | √3 | ||
cosx − | sinx = 0 | ||
2 | 2 |
π | π | |||
sin | cosx − cos | sinx = 0 | ||
6 | 6 |
π | ||
sin(x − | ) = 0 | |
6 |
π | ||
t = x − | ||
6 |
π | ||
x− | = kπ | |
6 |
π | π | |||
x = kπ + | −−−> Czyli wykres przesuwamy o | w prawo. | ||
6 | 6 |
π | ||
sin(x − | ) = 1 | |
6 |
π | 2 | |||
x− | = | π + 2kπ | ||
6 | 3 |
5 | ||
x = | π + 2kπ | |
6 |
cosx−√3*sinx | ||
powinno być f(x) = [ | ]*2 | |
2 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
Kliknij po więcej przykładów | |
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Twój nick | |