math

posted by .

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 -

    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 -

    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 -

    thanks for explaining #1 :)


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

  • math -

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

  • math -

    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 -

    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 -

    OHHH, I forgot my chart,
    so x = pi/3 and 5pi/3 :)

    thank you very much!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. math

    y= 2sin^2 x y=1- sinx find values of x inthe interval 0<x<360 if 2sin^x = 1-sinx this can be arranged into the quadratic. 2sin^2 x + sin x -1=0 (2sinx -1)(sinx + 1)=0 or sinx = 1/2 and sin x=-1
  2. math

    the problem is 2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is what i've done so far: 2cos^2x+sinx-1=0 2cos^2x-1+sinx=0 cos2x + sinx =0 1 - 2sin^2x + sinx = 0 -2sin^2x+sinx-1=0 …
  3. math

    tanx+secx=2cosx (sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 = 2(1-sin^2x) 2sin^2x + sinx-1=0 (2sinx+1)(sinx-1)=0 x=30 x=270 but if i plug 270 back into the original equation …
  4. Trigonometry

    4. Find the exact value for sin(x+y) if sinx=-4/5 and cos y = 15/17. Angles x and y are in the fourth quadrant. 5. Find the exact value for cos 165degrees using the half-angle identity. 1. Solve: 2 cos^2x - 3 cosx + 1 = 0 for 0 less …
  5. calculus

    Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the derivative
  6. Trig Help

    Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1-cos^2x)]/[sinx+1] =?
  7. Math (trigonometric identities)

    I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx …
  8. Math

    1) evaluate without a calculator: a)sin(3.14/4) b) cos(-3(3.14)/4) c) tan(4(3.14)/3) d) arccos(- square root of three/2) e) csctheata=2 2) verify the following identities: a) cotxcosx+sinx=cscx b)[(1+sinx)/ cosx] + [cosx/ (1+sinx)]= …
  9. Math Help

    Hello! Can someone please check and see if I did this right?
  10. Math

    I need help solving for all solutions for this problem: cos 2x+ sin x= 0 I substituted cos 2x for cos^2x-sin^2x So it became cos^2(x)-sin^2(x) +sinx=0 Then i did 1-sin^2(x)-sin^2(x)+sinx=0 = 1-2sin^2(x)+sinx=0 = sinx(-2sinx+1)=-1 What …

More Similar Questions