Find all exact solutions on the interval [0, 2π). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) sin^3(t) = sin(t)

Answer 1

First, 2 notes about notation:

1) sin³(t) = sin(t) means [sin(t)]³ = sin(t).

2) [0, 2π) includes 0 but does NOT include 2π.

sin³(t) = sin(t)

Simplify this by considering sin(t) as a simple variable, like x.

So, now we have x³ = x

x³ - x = 0

x(x² - 1) = 0

x(x+1)(x-1) = 0

x = 0, -1, and 1

So, sin(t) = 0, -1, and 1. So, now all that's left to do is to find the values of t for which sin(t) is 0, -1, or 1. You can do that by knowing the unit circle or by using inverse sin (e.g. sin^-1(1) = π/2).

On the interval [0, 2π), where does sin(t) = 0? That occurs when t = 0 or π (note: sin(t) = 0 when t = 2π but 2π isn't within the interval so it's not a solution to this problem).

On the interval [0, 2π), where does sin(t) = -1? That occurs when t = 3π/2

On the interval [0, 2π), where does sin(t) = 1? That occurs when t = π/2

So, putting all that together, the solutions are 0, π/2, π, and 3π/2.

Answer 2

Let M = sin t

M^3 = M

M^3 - M = 0

M(M^2 - 1) = 0

M(M+1)(M-1) = 0

M=0 M= -1 M = 1

sin = 0 sin = -1 sin = 1

so the base angles are 0, 90,180,270