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Homework Help: Please help I cant Seem to find these out

  1. Jan 9, 2008 #1
    Sorry, I don't know how to type in some of the symbols.

    1. Solve Each Equation For 0 (degrees)_< x _<180 (degrees)


    1. sin 2x = cos 3x
    2. 2cos^2x = sinx +1
    3. cos^2x - 7/2cosx-2=0
    4.tan^2x = 3tanx
    5. sin2x=cos3x

    3. The attempt at a solution

    1. sin2x - cos3x = 0
    2sinx - cos3x=0

    2. square both sides????

    3. cos^2x - 7/2cosx - 2 Factored??

    4.I'm sorry, no idea

    5. sin2x - cos3x = o

    Thanks a lot, I am having huge trouble with this subject. Any guides, and help would be very largely appreciated, I have a huge test on Friday. Thanks a lot
  2. jcsd
  3. Jan 9, 2008 #2
    I'm going to help you, but you're gonna have to work every single problem with me.

    1. [tex]\sin{2x}-\cos{3x}=0[/tex]

    What is you reasoning for [tex]\sin{2x} \rightarrow 2\sin{x}[/tex] Is that an identity?
  4. Jan 9, 2008 #3
    no I'm sorry, i must have typed something wrong, that is not an identity....I was making an attempt, and ended up deleting it because it was wrong....I mistakenly left that in.
  5. Jan 9, 2008 #4
    That's fine, let's start by doing a simple algebraic manipulation to cosine 3x.

    [tex]\sin 2x = \cos{(2x+x)}[/tex]

    We have an identity for [tex]\cos{(x+y)}[/tex] which says that it's just cosxcosy-sinxsiny, so let's apply that to our manipulated cosine.
    Last edited: Jan 9, 2008
  6. Jan 9, 2008 #5
    ok, so

    sin2x = cos2x cosx - sin2x sinx

    RHS: cos2x = 2cos^2x - 1 in identities so,

    2cos^2x - 1 ( cosx) - 2sinxcosx (sinx)

    =2cos^3x-cosx - 2sinx^2 cosx(sinx)?
    ?? I think I am doing this way wrong
  7. Jan 9, 2008 #6
    I did it differently, but you're on the right track

    [tex]\sin 2x=\cos x\cos{2x}-\sin x\sin{2x}[/tex]

    [tex]2\sin x \cos x=\cos x(1-2\sin^{2}x)-2\sin^{2}x\cos x[/tex]

    [tex]2\sin x \cos x=\cos x - 4\sin^{2}x\cos x[/tex]

    Follow me? Next step, put it all on 1 side and factor out a common term of cosine.
    Last edited: Jan 9, 2008
  8. Jan 9, 2008 #7
    Still there?
  9. Jan 9, 2008 #8
    yes sorry i was eating dinner
  10. Jan 9, 2008 #9
    Is ok, I'm eating dinner as well. Just reply back when you're done or still want help.
  11. Jan 9, 2008 #10
    so this would then go to cosx (2sinx-x) -cosx(4sin^2)=0 ?
  12. Jan 9, 2008 #11
    It should become ...

    [tex]\cos x (4\sin^2 x + 2\sin x -1)=0[/tex]

    You factored out incorrectly. So from here, set them individually equal to 0.

    [tex]\cos x =0[/tex] This should yield 2 solutions.

    [tex]4\sin^2 x + 2\sin x - 1=0[/tex] You will need to use the quadratic formula in order to solve for your roots.
  13. Jan 9, 2008 #12
    ok i see, so for cosx = 0, this would be 90 degrees.

    Using the quadratic equation, i came out with -2 plus or minus (2*square root of 5)/8....i am doing something way wrong, the quadriatic equation is -b plus or minus the (square root of b^2 - 4ac)/2a......i used the numbers from 4sin^2x +2sinx-1=0
  14. Jan 9, 2008 #13
    You did it correctly.

    Also, I forgot this is only from 0-180, so be careful with the values you find from the quadratic equation.

    Remember pi is approx 3.14, so anything bigger just discard.

    After reducing ... [tex]\sin x =\frac{-1\pm\sqrt 5}{4}[/tex]
  15. Jan 9, 2008 #14
    Thanks a lot, i seem to be having trouble factoring simple things for some reason, such as 2 sin^2x -1 = 0....any tips or tricks that you use?


    for equations don't have a number infront of the "a" place, I usually think of numbers that add to b, and multiply to c....any tips for equations that have numbers in front of the A? Sorry, I know this is very simple math, but I never quite learned this too strongly.
  16. Jan 9, 2008 #15
    [tex]2\sin^2 x-1=0[/tex]

    Just move 1 to the other side, divide by 2, take the square root of both sides.

    [tex]\sin x =\pm\frac{1}{\sqrt{2}}[/tex]

    So your solution is ...


    If there isn't a number in the place of the a, then the number is simply 1.
    Last edited: Jan 9, 2008
  17. Jan 9, 2008 #16
    so moving onto the 2nd problem, i am confused on what to do.
  18. Jan 9, 2008 #17
    [tex]2\cos^2 x=\sin x +1[/tex]

    Think of an identity that you can use so your equation is in terms of 1 identity.
  19. Jan 9, 2008 #18
    sinx^2 + cosx^2 = 1 maybe??

    i really am not sure to do to this problem,
    maybe 2cos^2x - sinx +1=o?

    the sin x isn't squared though, I am not sure.
  20. Jan 9, 2008 #19

    Doesn't mean you have to change the sine term.

    Can you manipulate the identity [tex]sin^2 x+\cos^2 x=1[/tex] so that you can change your problem?
  21. Jan 9, 2008 #20
    I know you can, im just lost exactly what to do, maybe then....cos^2x -1 = sinx? then what do you do? for the identitiy sinx^2 + cosx^2 = 1 if you squared the whole thing it would be sinx + cosx =1...this isn't the same, i know that is wrong...i honestly don't know what to do.
  22. Jan 9, 2008 #21
    Well basically, I wanted you to replace the cosine squared term. Use the identity sin^2 x + cos^2 x = 1 to replace the cos^2 x of your problem.
  23. Jan 9, 2008 #22
    ohh...i see...hm, so then

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

    sinx=-2 ( no solution because less than -1) and sinx=1 (which is 90 degrees)

    Is this correct? Thanks you just turned on a light bulb for me, even if this is wrong, I now see it differently.
  24. Jan 9, 2008 #23
    It was really really really close :-]

    [tex]\cos^2 x = 1-\sin^2 x[/tex]

    [tex]2(1-\sin^2 x)=\sin x +1[/tex]

    [tex]2\sin^2 x +\sin x -1=0[/tex]

    Factor it.
    Last edited: Jan 9, 2008
  25. Jan 9, 2008 #24
    ok, so
    sinx=1, 1/2,

    x=90 degrees, and 30 degrees.

    Still wrong?

    Thanks a lot for your help, you have been a huggggeee help.
  26. Jan 9, 2008 #25
    Oh sorry about saying to use the quad. formula, you just had to factor it which I think you did anyways.

    But, only one of your answer is correct. The other should have been -1.

    [tex](2\sin^2 x-1)(\sin x+1)=0[/tex]
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