Find all real numbers in the interval [0,2pi) that satisfy the equation. Sin2x+sinx=0
To solve the equation sin(2x) + sin(x) = 0, we can use the double angle identity for sine, which states that sin(2x) = 2sin(x)cos(x).
Therefore, the equation becomes:
2sin(x)cos(x) + sin(x) = 0
Factoring out sin(x), we get:
sin(x)(2cos(x) + 1) = 0
This equation will be satisfied if either sin(x) = 0 or 2cos(x) + 1 = 0.
1. When sin(x) = 0:
This occurs at x = 0, π, and 2π.
2. When 2cos(x) + 1 = 0:
2cos(x) = -1
cos(x) = -1/2
This happens at x = 2π/3 and 4π/3.
Therefore, the real numbers in the interval [0, 2π) that satisfy the equation are: 0, π, 2π/3, and 4π/3.