Question

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

Answer 1

Step-by-step explanation:

sin2x - sinx = 0

2sinxcosx - sinx = 0

(Double Angle Formula)

sinx(2cosx - 1) = 0

Either sinx = 0 or cosx = 1/2.

When sinx = 0,

x = 0 or x = π.

When cosx = 1/2,

x = π/3 or x = 5π/3.

Hence the solutions are

x = 0, x = π/3, x = π or x = 5π/3.