posted by .

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:



cos2x + sinx =0

1 - 2sin^2x + sinx = 0



I think i'm supposed to factor it next but i'm not sure?


would doing that step be correct?

  • math -

    although, after looking at this problem i'm still not sure why i had to change the original problem? I've just been following the notes that were given to me by my teacher and that's what she has done but why do i have to change the original problem it seems like just a bunch of extra work

  • math -

    replace cos^2x by 1-sin^2x

    2(1-sin^2x) + sinx -1 = 0

    2-2sin^2x +sinx -1 = 0


    2sin^2x - sinx - 1=0

    (2sinx + 1)(sinx -1)=0

    we will get

    sinx = -1/2 and 1

  • math -

    carrying on where Dena left off....

    sinx = -1/2 (angle in standard postition = 30º)
    x must be in quadrant III or IV
    x = 180+30 =210 or
    x = 360-30 = 330

    if sinx = 1
    x = 90

    so in degrees x = 90, 210 or 330

    or in radians
    x = pi/2, 7pi/6 or 11pi/6

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Trig

    prove the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2) then multiply that out 1-2CosX^2 + cos^4 - cosX^2 + 2cos^4 -cos^6 add that on the left to the cos^6, and …
  2. Trig

    Verify the identity: tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx cosx - sinx/cosx =(2sinxcos^2 x - sinx)/cosx =sinx(2cos^2 x -1)/cosx = L.S. Q.E.D.
  3. trig

    find the exact solutions 2cos^2x+3sinx=0 the way it stands, that is a "nasty" question. Are you sure the second term isn't 2sin(2x) ?
  4. Trig.......

    I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr Reiny …
  5. Trig........

    I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top and …
  6. 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 …
  7. math

    solve each equation for 0=/<x=/<2pi sin^2x + 5sinx + 6 = 0?
  8. 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] =?
  9. Precalculus

    Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π). 8. cos2x=cosx 10. 2cos^2x+cosx=cos2x Solve algebraically for exact solutions in the interval [0,2π). Use your grapher only to support your …
  10. Math Help

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

More Similar Questions