Find the solutions to the given equation on the interval [0,2π)?
cot^3x−cotx=0
.
cotx(cot^2x-1) = 0
cotx(cotx-1)(cotx+1) = 0
These are angles you should recognize.
can you explain?
To find the solutions to the equation cot^3x - cotx = 0 on the interval [0, 2π), we can follow these steps:
Step 1: Simplify the equation
The given equation cot^3x - cotx = 0 can be rewritten as cotx(cot^2x - 1) = 0. Remember that cot^2x = 1 + csc^2x. Substituting this result, we get cotx(1 + csc^2x - 1) = 0, which simplifies to cotx(csc^2x) = 0.
Step 2: Identify the values for which cotx = 0 and csc^2x = 0
For the equation to be true, either cotx = 0 or csc^2x = 0.
When cotx = 0, we know that x belongs to the set {π/2, 3π/2} since cotx = 0 when x = π/2 or x = 3π/2.
When csc^2x = 0, this means cscx = 0, which is true only when x = 0 or x = π. However, these values are not within the given interval [0, 2π), so we exclude them from the solutions.
Step 3: Combine the solutions
Therefore, the solutions to the equation cot^3x - cotx = 0 on the interval [0, 2π) are {π/2, 3π/2}.