Basia:
1.
cosx = −
12 ⇔ x=π−α
0 lub x=π+α
0 ⇔ x=
2π3+2kπ lub x=
4π3+2kπ
czyli brak drugiego rozwiązania
2.
lub
|x| = 2π+6kπ = (6k+2)π = 2(3k+1)π
lub
|x| = 4π+6kπ = (6k+4)π = 2(3k+2)π
x = 2π+6kπ = (6k+2)π = 2(3k+1)π
lub
x = −[2π+6kπ] = −(6k+2)π = −2(3k+1)π
lub
x = 4π+6kπ = (6k+4)π = 2(3k+2)π
lub
x = −[4π+6kπ] = −(6k+4)π = −2(3k+2)π