Решение во вложении.
2cos(3x+8)=√2 /:2
cos(3x+8)=(√2)/2
cos 45°=cos (π/4)=(√2)/2
a)3x+8=π/4 + 2kπ,k∈Z
3x=π/4-8+2kπ,k∈Z
x=π/12-8/3+2kπ/3, k∈Z
b)3x+8=7π/4+2kπ, k∈Z
3x=7π/4-8+2kπ,k∈Z
x=7π/12-8/3+2kπ/3 , k∈Z
Решение во вложении.
2cos(3x+8)=√2 /:2
cos(3x+8)=(√2)/2
cos 45°=cos (π/4)=(√2)/2
a)3x+8=π/4 + 2kπ,k∈Z
3x=π/4-8+2kπ,k∈Z
x=π/12-8/3+2kπ/3, k∈Z
b)3x+8=7π/4+2kπ, k∈Z
3x=7π/4-8+2kπ,k∈Z
x=7π/12-8/3+2kπ/3 , k∈Z