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Find (4sin2x-1)sinx=0 for the interval [0,2pi)

asked Aug 18, 2017 in Algebra by Kile

1 Answer


 it's given that  sin^2x = 1/4

sin x = 1/2, x = pi/6, 5pi/6


sin x = -1/2, x = 7pi/6, 11pi/6

So the answer is

x = pi/6, 5pi, 7pi.6, 11pi/6.
answered Aug 21, 2017 by quickwriter (2,510 points)