# math

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?

1. 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

posted by Damon
2. 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 p-pi/6 = 5 pi/6

posted by Damon
3. thanks for explaining #1 :)

whoops, I posted the question wrong,
its
2sin^2 + sinx -6 = 0

posted by sh
4. LOL
2 y^2 + y - 6 = 0
(2 y- 3)(y + 2) = 0
still no good
|sin x| >1 for both solutions

posted by Damon
5. 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

posted by sh
6. 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

posted by Damon
7. OHHH, I forgot my chart,
so x = pi/3 and 5pi/3 :)

thank you very much!

posted by sh

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