2sin^2x - 5sinx + 3 < 0
this is how i am trying to solve it.
1. first i factored it:
(2sinx - 1) (sinx - 3) < 0
2. then i set (2sinx - 1) and (sinx - 3) equal to zero:
2sinx - 1 = 0
sinx = 1/2
x = pi/6, 5pi/6
sinx - 3 = 0
sinx = 3 (no solution)
3.
<-------------|----------|-----------------------|-------->
0 pi/6 5pi/6
now i find which x's give a negative value in the intervals (-infinity, pi/6), (pi/6, 5pi/6), and (5pi/6, infinity)
is this how i'm supposed to do it?
this is how i am trying to solve it.
1. first i factored it:
(2sinx - 1) (sinx - 3) < 0
2. then i set (2sinx - 1) and (sinx - 3) equal to zero:
2sinx - 1 = 0
sinx = 1/2
x = pi/6, 5pi/6
sinx - 3 = 0
sinx = 3 (no solution)
3.
<-------------|----------|-----------------------|-------->
0 pi/6 5pi/6
now i find which x's give a negative value in the intervals (-infinity, pi/6), (pi/6, 5pi/6), and (5pi/6, infinity)
is this how i'm supposed to do it?