d. sin(3x)= √2 / 2
(3x) = π/4 + 2kπ
oder:
(3x) = (π-π/4) + 2kπ
Jetzt beide noch nach x auflösen:
(3x) = π/4 + 2kπ
x= π/12 + 2kπ/3
oder:
(3x) = (π-π/4) + 2kπ
x= π/4 + 2kπ/3
L={x| x= π/12 + 2kπ/3 oder x= π/4 + 2kπ/3, k∈ℤ}
e. sin(x) = cos(x) |:cos(x)
tan(x) = 1
x = π/4 + kπ, k∈ℤ
L={x| x= π/4 + kπ, k∈ℤ}
f) sin(x) + cos² (x) = -1
sin(x) + (1-sin^2 (x)) = -1
0= sin^2 (x) - sin (x) -2
Subst. sin(x) = u
0= u^2 - u - 2 |faktorisieren
=(u-2)(u+1)
u1 = 2
u2=-1
rücksubst.
sin(x) = 2 → L1={}
sin(x)= -1 ---> x = 3π/4 + 2kπ, k∈ℤ