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Thread: A trig. question

  1. #1
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    A trig. question

    For range $\displaystyle 0`<=A<=360`$, solve:
    $\displaystyle 3cos^2 2A=1+sin A$

    Much help is appreciated. Thanks
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  2. #2
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    Quote Originally Posted by smmxwell View Post
    For range $\displaystyle 0`<=A<=360`$, solve:
    $\displaystyle 3cos^2 2A=1+sin A$

    Much help is appreciated. Thanks
    Substitute $\displaystyle \cos^2 A = 1 - \sin^2 A$ and re-arrange into an equation that looks something like the one in this thread: http://www.mathhelpforum.com/math-he...sin-x-3-a.html

    That thread tells you what to do next.
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  3. #3
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    Yeah. I know about that. Just that it's sin^2 2A. Even if i use the double angle formula, I still can't get anywhere. I know about factoring a q.e. but I just can't rearrange it.
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  4. #4
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    Quote Originally Posted by smmxwell View Post
    Yeah. I know about that. Just that it's sin^2 2A. Even if i use the double angle formula, I still can't get anywhere. I know about factoring a q.e. but I just can't rearrange it.
    My mistake. I misread the question. Are you expected to get exact solutions?
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  5. #5
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    Yeah. The answer is A= 30, 150, 221 deg. 49', 318 deg. 11'.
    I did my own steps and arrived at
    $\displaystyle 3cos^2 2A=(cos^2 A)/(1-sin A)$

    Even though $\displaystyle cos^2 2A$ is definitely not equal to $\displaystyle cos^2 A$, still, if I eliminate them, I manage to find $\displaystyle sin A= -2/3$, which would yield the last two readings. So, any help?

    btw, I just came back from a long break, so I can't really think straight now. I found another question that needs help too.

    Solve $\displaystyle tan x + 2 sin x=0$ for range=1 circle.
    How do you go further from $\displaystyle sin x +sin 2x= 0$
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  6. #6
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    Quote Originally Posted by smmxwell View Post
    Yeah. The answer is A= 30, 150, 221 deg. 49', 318 deg. 11'.
    I did my own steps and arrived at
    $\displaystyle 3cos^2 2A=(cos^2 A)/(1-sin A)$

    Even though $\displaystyle cos^2 2A$ is definitely not equal to $\displaystyle cos^2 A$, still, if I eliminate them, I manage to find $\displaystyle sin A= -2/3$, which would yield the last two readings. So, any help?

    btw, I just came back from a long break, so I can't really think straight now. I found another question that needs help too.

    Solve $\displaystyle tan x + 2 sin x=0$ for range=1 circle.
    How do you go further from $\displaystyle sin x +sin 2x= 0$
    I will solve a question is near to your see how I solve it

    $\displaystyle \sqrt{3}tan x =\frac{sinx }{cosx}$

    $\displaystyle \sqrt{3} \frac{sinx}{cosx} + 2sinx =0 $

    $\displaystyle \sqrt{3}sinx + 2sinx cos x =0 $

    $\displaystyle sinx(\sqrt{3}+2cosx)=0 $

    $\displaystyle sinx=0......x=n\pi$

    $\displaystyle \sqrt{3}+2cosx=0......cosx=-\frac{\sqrt{3}}{2}......x=....$

    your's is like this
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  7. #7
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    I'm sorry but I don''t really follow. You mean you're giving me a similar question which has the same steps? Either way, I don't see how you get $\displaystyle
    \sqrt{3} \frac{sinx}{cosx} + 2sinx =0
    $ after $\displaystyle
    \sqrt{3}tan x =\frac{sinx }{cosx}
    $

    the 2 question I posed weren't related. they are different questions which I both don't know how to do.
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  8. #8
    MHF Contributor Amer's Avatar
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    Ok I do a mistake in my solution

    $\displaystyle {\color{red}\sqrt{3}tan x =\frac{sinx }{cosx}}$ I forgot to multiply the answer with $\displaystyle \sqrt{3} $

    $\displaystyle {\color{blue}\sqrt{3}tan x =\sqrt{3}\left(\frac{sinx }{cosx}\right)}$ the correct one


    I will solve two of them Step by Step so look


    $\displaystyle \sqrt{3}tan x + 2 sin x =0 $

    $\displaystyle \sqrt {3} \left(\frac{sinx}{cosx}\right) + 2 sinx =0 $

    $\displaystyle \sqrt{3} sinx + 2 sinx cosx =0 $

    $\displaystyle sinx ( \sqrt{3} + 2 cos x ) =0 $

    sinx =0 so x=n(pi)
    cosx = -square(3)/2

    your's

    $\displaystyle tanx + 2 sinx =0$

    $\displaystyle \frac{sinx}{cosx} + 2sinx =0 $

    $\displaystyle sinx + 2 sinx cosx =0 $

    $\displaystyle sinx (1+ 2cosx ) =0 $

    $\displaystyle sinx =0 ......$

    $\displaystyle 1+2cosx =0.....cosx=\frac{-1}{2} $

    it is ok or not
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  9. #9
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    Yes. I get the 2nd one. But I still don't understand the first one;

    $\displaystyle
    3cos^2 2A=1+sin A
    $
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  10. #10
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by smmxwell View Post
    Yes. I get the 2nd one. But I still don't understand the first one;

    $\displaystyle
    3cos^2 2A=1+sin A
    $
    I will not solve it completely I will give you hints just

    $\displaystyle 3cos^2(2A)=1+sinA$

    you should know that

    $\displaystyle cos(2A)=1-2sin^2(A)$ find the

    $\displaystyle (1-2sin^2(A))^{2}$

    then plug it in your equation you will have an equation

    you will have

    $\displaystyle 3(1-4sin^2(A)+4sin^4(A))=1+sin(A)$

    $\displaystyle 3-12sin^2(A)+12sin^4(A)=1+sin(A)$

    rearrange it and let t=sin(A) you will have

    $\displaystyle 12t^4-12t^2-t+2=0$ this for you
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  11. #11
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    Trig equations

    Hello smmxwell
    Quote Originally Posted by smmxwell View Post
    For range $\displaystyle 0`<=A<=360`$, solve:
    $\displaystyle 3cos^2 2A=1+sin A$

    Much help is appreciated. Thanks
    If the answers you gave are correct then there's a mistake in the question. None of them satisfies $\displaystyle 3\cos^2 2A=1+\sin A$. (Try it and see!) However:

    $\displaystyle 3\cos2A = 1+\sin A$

    $\displaystyle \Rightarrow 3(1-2\sin^2A)=1+\sin A$

    $\displaystyle \Rightarrow 6\sin^2A +\sin A -2 = 0$

    $\displaystyle \Rightarrow (2\sin A -1)(3\sin A +2)=0$

    $\displaystyle \sin A = 0.5, -0.6667$

    $\displaystyle A = 30^o, 150^o, 211.81^o, 318.19^o$

    Grandad
    Last edited by Grandad; Jun 7th 2009 at 11:42 PM. Reason: Typo
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  12. #12
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    Yeah thats the anwser. I was trying to type in degrees and minutes.
    $\displaystyle
    \sin A = 30^o, 150^o, 211.81^o, 318.19^o
    $

    @Amer. I don't think quartic eq. is in my syllabus. I'm sure there's go to be a more elegant and simpler way?

    @ Grandad,

    The steps are for
    $\displaystyle
    3\cos2A = 1+\sin A
    $

    Whereas, the question wants $\displaystyle
    3\cos^2 2A=1+\sin A
    $

    your answer is correct though..
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  13. #13
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by smmxwell View Post
    Yeah thats the anwser. I was trying to type in degrees and minutes.
    $\displaystyle
    \sin A = 30^o, 150^o, 211.81^o, 318.19^o
    $

    @Amer. I don't think quartic eq. is in my syllabus. I'm sure there's go to be a more elegant and simpler way?

    @ Grandad,

    The steps are for
    $\displaystyle
    3\cos2A = 1+\sin A
    $

    Whereas, the question wants $\displaystyle
    3\cos^2 2A=1+\sin A
    $

    your answer is correct though..
    so you gave me a wrong equation that is not my fault I solve

    $\displaystyle 3cos^2(2A)=1+sinA$ since you gave me it

    but you did not gave me

    $\displaystyle 3cos(2A)=1+sinA $

    in math usually there is more than one way to solve questions that refer to the solver of the problem
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  14. #14
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    No. The question is $\displaystyle
    3cos^2(2A)=1+sinA
    $ and only that.

    Ok. Now I understand Grandad's post. The question is wrong. Alright. That means the question intended should be $\displaystyle
    3\cos2A = 1+\sin A
    $. Thanks a lot guys. Lol. I just spent the whole day trying to solve a wrong question.
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