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Asked by Hayli

find the solutions of the equation that are in the interval [0,2pi)

2-cos^2x=4sin^2(1/2x)
14 years ago

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Answered by Reiny
2 - (1 - 2sin^2 (x/2) = 4sin^2 (x/2)
1 = 2 sin^2 (x/2)
sin^2 (x/2) = 1/2
sin (x/2) = 1/√2
x/2 = π/4 or x/2 = 3π/4

then x = π/2 or x = 3π/2
14 years ago
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find the solutions of the equation that are in the interval [0,2pi)

2-cos^2x=4sin^2(1/2x)

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