Question

Let sin(2x) – sin(x) = 0, where 0 ≤ x < 2π. What are the possible solutions for x?

A) x = 0, π
B) x = π/2, 3π/2
C) x = π/4, 5π/4
D) x = π/3, 2π/3

Answer 1

The possible solutions for the equation sin(2x) - sin(x) = 0 are x = 0, π, π/3, 5π/3.

Step-by-step explanation:

To solve the equation sin(2x) - sin(x) = 0, we can use the trigonometric identity sin(2x) = 2sin(x)cos(x). Substituting this into the equation gives 2sin(x)cos(x) - sin(x) = 0. Factoring out sin(x), we have sin(x)(2cos(x) - 1) = 0.

Setting sin(x) = 0, we find that x = 0 and x = π are solutions. For the second factor, 2cos(x) - 1 = 0, we find that cos(x) = 1/2. This occurs at x = π/3 and x = 5π/3.

Combining all the solutions, we have x = 0, π, π/3, 5π/3.