Find all values of x such that sin(2x) = sin(x) and 0 ≤ x ≤ 2π

Solution:

Given, sin(2x) = sin(x), 0 ≤ x ≤ 2π -----(1)

Using trigonometry formula,

sin(2x) = 2 sinx cosx

Substituting the above formula in equation 1, we get

2 sinx cosx = sinx

2 sinx cosx - sinx = 0

Taking out common term,

⇒ sin x (2 cos x - 1) = 0

Now, sin x = 0

⇒ x = sin^-1(0)

⇒ x = 0, π, 2π

Now, 2 cos x - 1 = 0

⇒ 2 cos x = 1

⇒ cos x = 1/2

⇒ x = cos^-1(1/2)

x = π/3, 5π/3.

Therefore, the solutions are x = 0, π, 2π, π/3, 5π/3.

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Find all values of x such that sin(2x) = sin(x) and 0 ≤ x ≤ 2π

Summary:

All values of x such that sin(2x) = sin(x) and 0 ≤ x ≤ 2π are x = 0, π, 2π, π/3, 5π/3.