Algebra
Calculus
Trigonometry
Matrix
Question
Solve the inequality
−2arcsec(4)+kπ<x<2arcsec(4)+kπ∪45∘+kπ<x≤2π+kπ∪2π+kπ≤x<135∘+kπ,k∈Z
Alternative Form
−37.761244∘+180∘k<x<37.761244∘+180∘k∪45∘+180∘k<x≤90∘+180∘k∪90∘+180∘k≤x<135∘+180∘k,k∈Z
Alternative Form
−2arcsec(4)+kπ<x<2arcsec(4)+kπ∪2π+kπ≤x<43π+kπ∪4π+kπ<x≤2π+kπ,k∈Z
Evaluate
sec(2x)<4
Find the domain
More Steps
Evaluate
2x=2π+kπ,k∈Z
Divide both sides
22x=22π+kπ,k∈Z
Divide the numbers
x=22π+kπ,k∈Z
Divide the numbers
x=4π+2kπ,k∈Z
sec(2x)<4,x=4π+2kπ,k∈Z
Solve the equation
More Steps
Evaluate
sec(2x)=4
Use the inverse trigonometric function
2x=arcsec(4)
Calculate
2x=−arcsec(4)2x=arcsec(4)
Add the period of 2kπ,k∈Z to find all solutions
2x=−arcsec(4)+2kπ,k∈Z2x=arcsec(4)+2kπ,k∈Z
Calculate
More Steps
Evaluate
2x=−arcsec(4)+2kπ
Divide both sides
22x=2−arcsec(4)+2kπ
Divide the numbers
x=2−arcsec(4)+2kπ
Divide the numbers
x=−2arcsec(4)+kπ
x=−2arcsec(4)+kπ,k∈Z2x=arcsec(4)+2kπ,k∈Z
Calculate
More Steps
Evaluate
2x=arcsec(4)+2kπ
Divide both sides
22x=2arcsec(4)+2kπ
Divide the numbers
x=2arcsec(4)+2kπ
Divide the numbers
x=2arcsec(4)+kπ
x=−2arcsec(4)+kπ,k∈Zx=2arcsec(4)+kπ,k∈Z
Find the union
x={−2arcsec(4)+kπ2arcsec(4)+kπ,k∈Z
x={−2arcsec(4)+kπ2arcsec(4)+kπ,k∈Z
Find the monotonically intervals
[kπ,45∘+kπ),k∈Z(45∘+kπ,2π+kπ],k∈Z[2π+kπ,135∘+kπ),k∈Z(135∘+kπ,π+kπ],k∈Z
Find the intersection
More Steps
Evaluate
[kπ,45∘+kπ),k∈Z
Determine the boundary values associated with the monotonic intervals
[kπ,45∘+kπ),k∈Z2arcsec(4)+kπ
Given that sec(2x) is monotonically increasing on [kπ,45∘+kπ),k∈Z and the inequality is sec(2x)<4, we conclude that x<2arcsec(4)+kπ,k∈Z
{kπ≤x<45∘+kπ,k∈Zx<2arcsec(4)+kπ,k∈Z
Find the intersection
kπ≤x<2arcsec(4)+kπ,k∈Z
kπ≤x<2arcsec(4)+kπ,k∈Z(45∘+kπ,2π+kπ],k∈Z[2π+kπ,135∘+kπ),k∈Z(135∘+kπ,π+kπ],k∈Z
Find the intersection
More Steps
Evaluate
(45∘+kπ,2π+kπ],k∈Z
Find the maximum value of sec(2x) on the interval (45∘+kπ,2π+kπ],k∈Z
The maximum value is −1
Given −1<4, we conclude that the solution set is 45∘+kπ<x≤2π+kπ,k∈Z
45∘+kπ<x≤2π+kπ,k∈Z
kπ≤x<2arcsec(4)+kπ,k∈Z45∘+kπ<x≤2π+kπ,k∈Z[2π+kπ,135∘+kπ),k∈Z(135∘+kπ,π+kπ],k∈Z
Find the intersection
More Steps
Evaluate
[2π+kπ,135∘+kπ),k∈Z
Find the maximum value of sec(2x) on the interval [2π+kπ,135∘+kπ),k∈Z
The maximum value is −1
Given −1<4, we conclude that the solution set is 2π+kπ≤x<135∘+kπ,k∈Z
2π+kπ≤x<135∘+kπ,k∈Z
kπ≤x<2arcsec(4)+kπ,k∈Z45∘+kπ<x≤2π+kπ,k∈Z2π+kπ≤x<135∘+kπ,k∈Z(135∘+kπ,π+kπ],k∈Z
Find the intersection
More Steps
Evaluate
(135∘+kπ,π+kπ],k∈Z
Modify the boundary values based on the period
(135∘+kπ,π+kπ],k∈Z−2arcsec(4)+π
Determine the boundary values associated with the monotonic intervals
(135∘+kπ,π+kπ],k∈Z2−arcsec(4)+2π
Given that sec(2x) is monotonically decreasing on (135∘+kπ,π+kπ],k∈Z and the inequality is sec(2x)<4, we conclude that x>2−arcsec(4)+2π+kπ,k∈Z
{135∘+kπ<x≤π+kπ,k∈Zx>2−arcsec(4)+2π+kπ,k∈Z
Find the intersection
2−arcsec(4)+2π+kπ<x≤π+kπ,k∈Z
kπ≤x<2arcsec(4)+kπ,k∈Z45∘+kπ<x≤2π+kπ,k∈Z2π+kπ≤x<135∘+kπ,k∈Z2−arcsec(4)+2π+kπ<x≤π+kπ,k∈Z
Calculate
−2arcsec(4)+kπ<x<2arcsec(4)+kπ∪45∘+kπ<x≤2π+kπ∪2π+kπ≤x<135∘+kπ,k∈Z
Check if the solution is in the defined range
−2arcsec(4)+kπ<x<2arcsec(4)+kπ∪45∘+kπ<x≤2π+kπ∪2π+kπ≤x<135∘+kπ,k∈Z,x=4π+2kπ,k∈Z
Solution
−2arcsec(4)+kπ<x<2arcsec(4)+kπ∪45∘+kπ<x≤2π+kπ∪2π+kπ≤x<135∘+kπ,k∈Z
Alternative Form
−37.761244∘+180∘k<x<37.761244∘+180∘k∪45∘+180∘k<x≤90∘+180∘k∪90∘+180∘k≤x<135∘+180∘k,k∈Z
Alternative Form
−2arcsec(4)+kπ<x<2arcsec(4)+kπ∪2π+kπ≤x<43π+kπ∪4π+kπ<x≤2π+kπ,k∈Z
Show Solution
