Question

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`

Answer 1

`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]}