math
posted by sh .
solve each equation for 0=/<x=/<2pi
sin^2x + 5sinx + 6 = 0?
how do i factor this and solve?
2sin^2 + sinx = 0
(2sinx  3)(sinx +2)
sinx = 3/2, 2
how do I solve? These are not part of the special triangles.
2cos^2  7cosx + 3 = 0
(2cosx  1)(cosx  3)
cosx = 1/2, 3
x= pi/4, 7pi/4,
how do I find solve for 3?
thanks in advance

sin^2x + 5sinx + 6 = 0?
how do i factor this and solve?
by saying y = sin x
then you have
y^2 + 5 y + 6 = 0
(y+2)(y+3) = 0
y = 2 or y = 3
WHICH IS IMPOSSIBLE because sin x has to be 1 </= sin x </= +1 
2sin^2 + sinx = 0
You factored wrong
sin x (2 sin x +1) = 0
sin x = 0 which is at 0 and pi
sin x = 1/2
which is at pi/6 and at ppi/6 = 5 pi/6 
thanks for explaining #1 :)
whoops, I posted the question wrong,
its
2sin^2 + sinx 6 = 0 
LOL
2 y^2 + y  6 = 0
(2 y 3)(y + 2) = 0
still no good
sin x >1 for both solutions 
yeah, right after I read understood the first no solution, i figured the second was no solution =]
my answers for number 3 are wrong, but I don't understand why 
2cos^2  7cosx + 3 = 0
(2cosx  1)(cosx  3)
cosx = 1/2, 3
ok so far BUT
cos 60 degrees = 1/2
that is pi/3
NOT pi/4 
OHHH, I forgot my chart,
so x = pi/3 and 5pi/3 :)
thank you very much!
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