pre cal 2
posted by kc .
find the exact solutions in the interval [0,2pi), sin4x=2sin2x

sin4x=2sin2x
2 sin2xcos2x = 2sin2x, ......... since sin2A = 2sinAcosA
cos2x = 1
2x = 180°, 360°, 540°, 720° which is π , 2π 3π, or 4π radians
x = 90° ,180°,270°, 360° which is π/2, π , 3π/2, or 2π radians
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