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Mathematics 17 Online
OpenStudy (anonymous):

Please help. Solve the equation for exact solutions in the interval 0 ≤ x < 2π. sin4x - √2 sin2x = 0

OpenStudy (anonymous):

sin4x = sin2(2x) = 2sin2x cos2x

OpenStudy (anonymous):

sin4x - √2 sin2x = 0 sin4x = √2 sin2x

OpenStudy (anonymous):

2sin2x cos2x = √2 sin2x cos2x = 1/√2

OpenStudy (anonymous):

hope u can do the rest..:)

OpenStudy (anonymous):

having difficulties, maybe you could find x for me?

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