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Solve for [0,2pi)

2cos4x + sqrt3 = 0

1 answer

2cos4x + √3 = 0
cos4x = -√3/2
4x = 5π/6 or 7π/6
x = 5π/24 or 7π/24
period is 2π/4 = π/2, so all solutions are

5π/24 + kπ/2 or 7π/24 + kπ/2

So, for all solutions in [0,2π) we have

5π/24, 17π/24, 29π/24, 41π/24
7π/24, 19π/24, 31π/24, 43π/24
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