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?

thanks in advance

asked by sh
  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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