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# Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0

**Solution:**

Given, sin 2x - sin 4x = 0

sin (2x) = sin (4x)

sin (2x) = sin (2 × 2x)

Sin (2x) = 2 sin (2x) cos (2x)

Subtracting sin (2x) on both sides,

2 sin (2x) cos (2x) - sin (2x) = 0

Taking sin (2x) as common,

sin (2x) [(2cos (2x) -1)] = 0

Now using product rule,

sin (2x) = 0

2cos (2x) -1 = 0

cos (2x) = 1/2

This implies x = 0+(nπ/2), π/6 + nπ, 5π/6 + nπ

x = 0, π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6, 2π.

Therefore, the solutions to the equation is 0,π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6, 2π.

## Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0

**Summary:**

All solutions to the equation sin 2x - sin 4x = 0 in the interval [0, 2π) is x = 0,π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6, 2π.

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