Asked by SHINee

Solve sin^2x - 2sinx - 3 = 0 on the interval xE [0, 2pie]
Answered by Steve
pi, not pie

sin^2x - 2sinx - 3 = 0
(sinx-3)(sinx+1) = 0
sinx=3 nope
sinx = -1
x = 3pi/2
Answered by Sara
4cos2x - cosx-3 =0

0 to 2pi
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