Question
Solve the inequality
45π+2kπ<x<47π+2kπ,k∈Z
Alternative Form
x∈(45π+2kπ,47π+2kπ),k∈Z
Alternative Form
225∘+360∘k<x<315∘+360∘k,k∈Z
Evaluate
sin(x)<−22
Solve the equation
More Steps

Evaluate
sin(x)=−22
Use the inverse trigonometric function
x=arcsin(−22)
Calculate
x=45πx=47π
Add the period of 2kπ,k∈Z to find all solutions
x=45π+2kπ,k∈Zx=47π+2kπ,k∈Z
Find the union
x={45π+2kπ47π+2kπ,k∈Z
x={45π+2kπ47π+2kπ,k∈Z
Find the monotonically intervals
[2π+2kπ,23π+2kπ],k∈Z[23π+2kπ,25π+2kπ],k∈Z
Find the intersection
More Steps

Evaluate
[2π+2kπ,23π+2kπ],k∈Z
Determine the boundary values associated with the monotonic intervals
[2π+2kπ,23π+2kπ],k∈Z45π+2kπ
Given that sin(x) is monotonically decreasing on [2π+2kπ,23π+2kπ],k∈Z and the inequality is sin(x)<−22, we conclude that x>45π+2kπ,k∈Z
{2π+2kπ≤x≤23π+2kπ,k∈Zx>45π+2kπ,k∈Z
Find the intersection
45π+2kπ<x≤23π+2kπ,k∈Z
45π+2kπ<x≤23π+2kπ,k∈Z[23π+2kπ,25π+2kπ],k∈Z
Find the intersection
More Steps

Evaluate
[23π+2kπ,25π+2kπ],k∈Z
Determine the boundary values associated with the monotonic intervals
[23π+2kπ,25π+2kπ],k∈Z47π+2kπ
Given that sin(x) is monotonically increasing on [23π+2kπ,25π+2kπ],k∈Z and the inequality is sin(x)<−22, we conclude that x<47π+2kπ,k∈Z
{23π+2kπ≤x≤25π+2kπ,k∈Zx<47π+2kπ,k∈Z
Find the intersection
23π+2kπ≤x<47π+2kπ,k∈Z
45π+2kπ<x≤23π+2kπ,k∈Z23π+2kπ≤x<47π+2kπ,k∈Z
Solution
45π+2kπ<x<47π+2kπ,k∈Z
Alternative Form
x∈(45π+2kπ,47π+2kπ),k∈Z
Alternative Form
225∘+360∘k<x<315∘+360∘k,k∈Z
Show Solution
