Question

Решить уравнения:
1) sin(3pi/2-x)=1/3
2) cos(2x-pi/3)=-1

Answer 1

1) sin(3π/2-x)=1/3
3π/2-x=(-1)^k * arcsin(1/3)+πk, k∈Z
-x=(-1)^k * arcsin(1/3)-3π/2+πk, k∈Z
x=(-1)^k+1 * arcsin(1/3)+3π/2-πk, k∈Z

2) cos(2x-π/3)=-1
2x-π/3=π-2πk, k∈Z
2x=π+π/3-2πk, k∈Z
x=2π/3-πk, k∈Z