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

**Solution:**

Given, cos 4x - cos 2x = 0 -----------------------(1)

From trigonometric identities,

cos^{2}x + sin^{2}x = 1 and cos 2x = cos^{2}x - sin^{2}x

Hence, we can write cos 4x as:

cos(4x) = cos^{2}(2x) - sin^{2}(2x) = 2cos^{2}(2x) -1 ----------(2)

Substitute (2) in (1)

2cos^{2}(2x) -1 - cos2x = 0 ---------------------(3)

Now, consider cos 2x = t

(3) becomes 2t^{2} - 1 - t = 0

On solving,

2t^{2} - 2t + t - 1= 0

2t(t - 1) + 1(t - 1) = 0

(2t + 1)(t - 1) = 0

2t + 1 = 0

t = -1/2

t - 1 = 0

t = 1

This implies cos 2x = -1/2 or 1.

Given the interval is [0, 2π].

cos 2x = 1

x = 0

cos (2π - 2x) = 1

cos (2π - 0) = 1

x = π

Cos 2x = -1/2

2x = 2π/3 or 2x = 4𝜋/3

x = π/3 or 2π/3

cos (2π - 2x) = -1/2

2π - 2x = 2π/3 or 2π - 2x = 4π/3

x = 2π/3 or 4π/3

Therefore, the solutions are x = 0, π/3, 2π/3, 4π/3, 5π/3, π.

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

**Summary:**

All solutions to the equation cos 4x - cos 2x = 0 in the interval [0, 2π] are x = 0, π/3, 2π/3, 4π/3, 5π/3, π.

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