1. cos (x/3) = -1
x/3 = π + 2πk, k∈Z
2. ctg ((2x)/7) = 1
ctg ((2x)/7) = 1, x≠((7πk)/2), k∈Z
(2x)/7 = arcctg(1)
(2x)/7 = π/4 + πk, k∈Z
8x = 7π + 28πk, k∈Z
x = (7π)/8 + (7πk)/2, k∈Z
3) tg(7x + π/6) = 4,1
tg(7x + π/6) = 4,1, x≠π/21 + πk/7, k∈Z
7x = arctg(4,1) - π/6 +πk, k∈Z
x = (arctg(4,1))/7 - π/42 + πk/7, k∈Z
4) 2cos ((x/11) - (π/6)) + √3 = 0
2cos ((x/11) - (π/6)) = -√3
cos ((x/11) - (π/6))= -(√3/2)
ответ: x=11π + 22πk, k∈Z; x=(44π)/3 + 22πk, k∈Z
5) -2sin(4x - π/9) = -0,4
4x - π/9 = arcsin(1/5)
sin(10π/9 - 4x)=1/5
ответ: x= (arcsin(1/5))/4 + π/36 + πk/2, k∈Z; (arcsin(1/5))/4 + 5π/18 + πk/2, k∈Z
6) 3cos x - 2 sin x = 0
2sin x = 3cos x | /cos x
2tg x = 3
tg x = 3/2
x = arctg(3/2) + πk, k∈Z
7) 4cos²(x/5) -3 = 0
4cos²(x/5) = 3
cos²(x/5) = 3/4
cos(x/5) = ±((√3)/2)
ответ: x = 5π/6 + 5πk, k∈Z; x = 25π/6 + 5πk, k∈Z
8) 3sin x - 2 cos x = 3
3 * (2t/(1+t²)) - 2((1-t²)/(1+t²)) = 3
t=1
t=5
tg(x/2) = 1
tg(x/2) = 5
x = π/2 + kπ, k∈Z; x=2arctg5 + 2kπ, k∈Z
Пошаговое объяснение:
