Find all values of x such that sin 2x = sin x and 0 x 2π. (List the answers in increasing order.)

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To find all values of x such that sin 2x = sin x, we can use the trigonometric identity sin 2x = 2sin x cos x.

So, we have 2sin x cos x = sin x.

Let's consider two cases:

Case 1: sin x = 0
If sin x = 0, then x can be 0, π, or 2π, since these are the values in the given range where sin x = 0.

Case 2: sin x ≠ 0
In this case, we can divide both sides of the equation by sin x:
2cos x = 1

Now, let's solve for x by considering the range 0 ≤ x ≤ 2π:

If cos x = 1/2, we know that this occurs at two angles in the range 0 ≤ x ≤ 2π: π/3 and 5π/3.

So, combining all the solutions, we have:

x = 0, π, 2π, π/3, and 5π/3.

Listing the answers in increasing order, we get:

x = 0, π/3, 5π/3, π, 2π.

To find all the values of x that satisfy the equation sin 2x = sin x, we can use the fact that sin 2x = 2sin x cos x.

Therefore, the equation can be rewritten as 2sin x cos x = sin x.

First, let's solve for sin x = 0. Since we know that sin 0 = 0 and sin π = 0, we have two solutions: x = 0 and x = π.

Next, let's solve for cos x in the equation 2sin x cos x = sin x. We can divide both sides of the equation by sin x (assuming sin x ≠ 0) to get 2cos x = 1. This implies that cos x = 1/2.

The values of x that satisfy cos x = 1/2 are x = π/3 and x = 5π/3.

Finally, let's combine all the solutions, x = 0, π, π/3, and 5π/3, in increasing order:

x = 0, π/3, π, 5π/3.

sin 2x - sinx = 0 , 0 < x < 2π

2sinx cosx - sinx = 0
sinx(2cosx - 1) = 0
sinx = 0 or cosx = 1/2

for sinx = 0
x = 0 , π , 1π

for cosx = 1/2
x = π/3 or x = 5π/3