Let n be a fixed positive integer. Describe all solutions of sin(n theta) = 1/2
I know that the solutions would be
(π/6)/n+(2πx)/n
and
(5π/6)/n+(2πx)/n
?
How do I 'describe'
7 years ago
7 years ago
the period is 2π/n
so, the solutions are
π/6n + k(2π/n) = π/6n (12k+1)
5π/6n + k(2π/n) = π/6n (12k+5)
for all integers k.
11 months ago
To describe all the solutions of the equation sin(nθ) = 1/2, where n is a fixed positive integer, you can use the general solution for the sine function.
For any value of θ, the sin function produces a value between -1 and 1. When sin(nθ) = 1/2, it means that the angle nθ has a corresponding reference angle that gives a result of 1/2.
To find the solutions, you can consider the inverse sine function, often denoted as arcsin or sin^(-1). The inverse sine function takes a value between -1 and 1 and returns an angle between -π/2 and π/2.
The general solution for sin(nθ) = 1/2 can be written as:
nθ = π/6 + 2πk or nθ = 5π/6 + 2πk
where k is an integer.
To isolate θ, divide both sides of the equations by n:
θ = π/(6n) + (2πk)/n or θ = 5π/(6n) + (2πk)/n
So, the solutions are given by:
θ = (π/6)/n + (2πk)/n or θ = (5π/6)/n + (2πk)/n
where k is an integer.
This describes all the solutions of the equation sin(nθ) = 1/2 for a fixed positive integer n.