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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: destain on November 02, 2012, 07:12:10 pm

Title: General Solutions
Post by: destain on November 02, 2012, 07:12:10 pm: Let h: R --> R, h(x) = 2 |sin(x)|. Find the general solution for x of the equation h(x) = 1.
Its CAS active so I just solved using my CAS and it didn't give me the right answer :S?..
I remembered the abs and everything
Title: Re: General Solutions
Post by: Jenny_2108 on November 02, 2012, 07:15:37 pm: works fine in my CAS. Do you use TI-nspire?
Title: Re: General Solutions
Post by: destain on November 02, 2012, 07:29:19 pm: yup and It gives me 3 solutions but in the answers, it's something else i'm pretty sure?
And how would you do it manually anyway also it's 2009 VCAA 1d in section B Paper 2
Title: Re: General Solutions
Post by: BigAl on November 02, 2012, 07:33:49 pm: You take 2 down there . so abs(sin(x))=1/2
To get rid off abs you put +- on the other side ...and solve those 2 equations.
Title: Re: General Solutions
Post by: destain on November 02, 2012, 07:39:56 pm: Yeah that's just rearranging it isn't it? Still the same solution, do you get the correct ones when you solve it on the CAS? One of the three is correct that I get
Title: Re: General Solutions
Post by: polar on November 02, 2012, 07:40:09 pm: to do this by hand,
$\begin{aligned} 2|\sin(x)| &= 1 \\ |\sin(x)| &= \frac{1}{2} \\ \sin(x) &= \pm\frac{1}{2} \\ \sin(x) &= \frac{1}{2} \; \text{or} \; -\frac{1}{2} \\ x &= 2\pi{n} + \frac{\pi}{6}, 2\pi{n} + \frac{5\pi}{6} \; \text{or} \; 2\pi{n} - \frac{\pi}{6}, 2\pi{n} - \frac{5\pi}{6}, \; n \in Z \end{aligned}$

these values are basically $\frac{\pi}{6}$ from both $\pi$ and $-\pi$ therefore, this can be simplified to $\pi{n} \pm \frac{\pi}{6}, \; n \in Z$
Title: Re: General Solutions
Post by: BigAl on November 02, 2012, 07:45:49 pm: I'm using casio classpad and..with algebraic expression my calculator gives me weird things. But you can also use graphing method..Just sketch y=1 and that function and see where they intersect. But be careful..set your calculator on radian mode..now you should get something very irrational number...I mean there could be numerical approximation but you have to express that in radian form..dont worry just use the ratio..if pi is 1 radian what would that number equal to in terms of radian? Sorry I can't show the representation here
Title: Re: General Solutions
Post by: destain on November 02, 2012, 07:51:56 pm: Quote from: polar on November 02, 2012, 07:40:09 pm
to do this by hand,
$\begin{aligned} 2|\sin(x)| &= 1 \\ |\sin(x)| &= \frac{1}{2} \\ \sin(x) &= \pm\frac{1}{2} \\ \sin(x) &= \frac{1}{2} \; \text{or} \; -\frac{1}{2} \\ x &= 2\pi{n} + \frac{\pi}{6}, 2\pi{n} + \frac{5\pi}{6} \; \text{or} \; 2\pi{n} - \frac{\pi}{6}, 2\pi{n} - \frac{5\pi}{6}, \; n \in Z \end{aligned}$

these values are basically $\frac{\pi}{6}$ from both $\pi$ and $-\pi$ therefore, this can be simplified to $\pi{n} \pm \frac{\pi}{6}, \; n \in Z$

That's right...How exactly do you simplify that down to your last answer?
I probably have the right answer from the CAS, just don't know how to simplify it to what they have here...?
They gave me 2npi - pi/6
2npi + 7pi/6
2npi + 5pi/6
Title: Re: General Solutions
Post by: BigAl on November 02, 2012, 07:55:58 pm: In this case the solution is repeating itself in pi period.
Title: Re: General Solutions
Post by: destain on November 02, 2012, 08:03:49 pm: I have no clue about how these equations got simplified ;( and also what my answer on the CAS is
Title: Re: General Solutions
Post by: polar on November 02, 2012, 08:20:45 pm: consider two of the solutions
$\begin{aligned}x &= 2\pi{n} \pm \frac{5\pi}{6} \\ &= (\pi{n} \pm ({\pi - \frac{\pi}{6}})) + \pi{n} \\ &= (\pi{n} \pm \pi) \pm \frac{\pi}{6} + \pi{n} \\ &= (2\pi{n} \pm \pi}) \pm \frac{\pi}{6} \end{aligned}$

for an integer, n, the value of 2n-1 and 2n+1 is always odd, therefore, let k=2n+1
$x = \pi{k_1} \pm \frac{\pi}{6}$

Now, consider the other two solutions
$x = 2\pi{n} \pm \frac{\pi}{6}$

clearly, for an integer, n, 2n is always even. therefore, you have both even and odd multiplies of $\pi$
$x = \pi{k_2} \pm \frac{\pi}{6}$

thus, you can simplify this to $x = \pi{k} \pm \frac{\pi}{6}$ since all integers are represented as k
Title: Re: General Solutions
Post by: Homer on November 02, 2012, 08:23:12 pm: I would just get sin(x) =1/2 graph and get all the x-values for which y = (1/2) and (-1/2) since its an absolute function
Title: Re: General Solutions
Post by: destain on November 02, 2012, 08:43:31 pm: Why does the CAS give the answers
2 nPi - pi/6
2 nPi + 7pi/6
2 nPi + 5pi/6

And the simplying is just confusing me at the moment..Don't think it's required for the mark anyway...T_T Methods is so frustrating haha
Title: Re: General Solutions
Post by: polar on November 02, 2012, 08:53:06 pm: Quote from: destain on November 02, 2012, 08:43:31 pm
Why does the CAS give the answers
2 nPi - pi/6
2 nPi + 7pi/6
2 nPi + 5pi/6

And the simplying is just confusing me at the moment..Don't think it's required for the mark anyway...T_T Methods is so frustrating haha

was $2\pi{n} + \frac{\pi}{6}$ also a solution? because that would be $\frac{\pi}{6}$ reflected into every quadrant which makes sense because it had the modulus signs at the start. I don't think the working was needed either :\
Title: Re: General Solutions
Post by: destain on November 02, 2012, 08:58:48 pm: Oh yes it is another solution..Wow this question is just terrible haha, probably really simple as well, I just don't get it.. T_T

and...another question...

Find the values of k, where k is a positive real number, for whic the equation 3 - ke^x - e^-x = 0 has one more solutions for x
Title: Re: General Solutions
Post by: Homer on November 02, 2012, 09:17:04 pm: has one more solutions for x?
Title: Re: General Solutions
Post by: Jenny_2108 on November 02, 2012, 09:30:17 pm: Quote from: destain on November 02, 2012, 08:58:48 pm
Oh yes it is another solution..Wow this question is just terrible haha, probably really simple as well, I just don't get it.. T_T

and...another question...

Find the values of k, where k is a positive real number, for whic the equation 3 - ke^x - e^-x = 0 has one more solutions for x

3 - ke^x - e^-x = 0
=> 3e^x-ke^(2x) -1=0
Let e^x=a
Solve 3a-ka^2-1=0 by using quadratic formula

Do you mean one or more solution or more than one solution?
If one or more sol, let delta>=0, if more than one solution, delta>0
Title: Re: General Solutions
Post by: destain on November 02, 2012, 09:45:17 pm: Quote from: Ennjy on November 02, 2012, 09:30:17 pm
Quote from: destain on November 02, 2012, 08:58:48 pm
Oh yes it is another solution..Wow this question is just terrible haha, probably really simple as well, I just don't get it.. T_T

and...another question...

Find the values of k, where k is a positive real number, for whic the equation 3 - ke^x - e^-x = 0 has one more solutions for x

3 - ke^x - e^-x = 0
=> 3e^x-ke^(2x) -1=0
Let e^x=a
Solve 3a-ka^2-1=0 by using quadratic formula

Do you mean one or more solution or more than one solution?
If one or more sol, let delta>=0, if more than one solution, delta>0

Sorry, I meant one or more, so do you have to solve using the quadractic formula? Or do you just use the discriminant thing? And also in this case would it be...a = 3 b= -k c= -1
or do you leave out the signs
Title: Re: General Solutions
Post by: Jenny_2108 on November 02, 2012, 11:54:17 pm: Quote from: destain on November 02, 2012, 09:45:17 pm
Quote from: Ennjy on November 02, 2012, 09:30:17 pm
Quote from: destain on November 02, 2012, 08:58:48 pm
Oh yes it is another solution..Wow this question is just terrible haha, probably really simple as well, I just don't get it.. T_T

and...another question...

Find the values of k, where k is a positive real number, for whic the equation 3 - ke^x - e^-x = 0 has one more solutions for x

3 - ke^x - e^-x = 0
=> 3e^x-ke^(2x) -1=0
Let e^x=a
Solve 3a-ka^2-1=0 by using quadratic formula

Do you mean one or more solution or more than one solution?
If one or more sol, let delta>=0, if more than one solution, delta>0

Sorry, I meant one or more, so do you have to solve using the quadractic formula? Or do you just use the discriminant thing? And also in this case would it be...a = 3 b= -k c= -1
or do you leave out the signs

You use discriminant>=0 only because they ask about value of k so no need to find x value I think
Discriminant= 9-4(-k)(-1) >=0 then solve for k