Posted by **sh** on Sunday, February 15, 2009 at 7:54pm.

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

- math -
**Damon**, Sunday, February 15, 2009 at 7:58pm
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

- math -
**Damon**, Sunday, February 15, 2009 at 8:00pm
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

- math -
**sh**, Sunday, February 15, 2009 at 8:02pm
thanks for explaining #1 :)

whoops, I posted the question wrong,

its

2sin^2 + sinx -6 = 0

- math -
**Damon**, Sunday, February 15, 2009 at 8:09pm
LOL

2 y^2 + y - 6 = 0

(2 y- 3)(y + 2) = 0

still no good

|sin x| >1 for both solutions

- math -
**sh**, Sunday, February 15, 2009 at 8:11pm
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

- math -
**Damon**, Sunday, February 15, 2009 at 8:14pm
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

- math -
**sh**, Sunday, February 15, 2009 at 8:16pm
OHHH, I forgot my chart,

so x = pi/3 and 5pi/3 :)

thank you very much!

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