How do you solve #sin^2x+1/2sinx > 0# ?

#sin^2x+1/2sinx > 0#
#t=sinx#
#t^2+1/2t>0 => t(t+1/2)>0#
Interval method
#t=0#
#t=1/2#
So
# - oo < sinx < -1/2 #
# 0< sinx< +oo #

But what do these inequalities mean (with #oo#)?
#sinx< +oo#
#sinx > -oo#

1 Answer
A. S. Adikesavan
Jul 30, 2018

#x notin ....U [ 13/6pi, -2pi ] U [ - pi, - 5/6pi ] U [ - 1/6pi, 0 ] U [ pi, 5/6pi ] U [11/6pi, 2pi ] U...#
See graphs and details. I would add more details, later.

Explanation:

#y = sin^2 x + 1/2 sin x = sin x ( sin x + 1/2 ) > 0#

#rArr# both #sin x and ( sin x + 1/2 )# have the same sign.

If sin x > 0, so is ( sin x + 1/2 ).

If sin x < 0, sin x < - 1/2, to make sin x + 1/2 < 0.

So, #sin x > 0 and sin x < - 1/2 #

This happens, when

#x notin ....U [ 13/6pi, -2pi ] u [ - pi, - 5/6pi ] U [ - 1/6pi, 0 ] U [ pi, 5/6pi ] U [11/6pi, 2pi ] U...#

See graph, for all y, with y-negative in #Q_3 and Q_4#.:
graph{ (y - 1/2 sin x - ( sin x )^2)= 0[-4 4 -2 2]}
x-not-to-be graph:
graph{ (y - 1/2 sin x - ( sin x )^2)(x+13/6pi+0.0001y)(x+2pi+0.0001y)(x+pi/6+0.0001y)(x+5pi/6+0.0001y)(x+pi+0.0001y)(x-pi+0.0001y)(x-7/6 pi+0.0001y)(x-2pi+0.0001y)(x-11/6pi+0.0001y)= 0[-8 8 -1 0]}

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