How is this incorrect?
- Thread starter Loki123
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Harry_the_cat
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Can you post the exact question please? Does it say "Solve the equation"?
We don't plug in our answers so I don't know how to that in trigonometry. It does state what x equals i just left that part out becaus I am more interested in why we got different answers.
D
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I did. It seems correct to me. Is it not?
Harry_the_cat
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When you said "The real result is: 2sin(3x/2)*2cosx*cos(x/2)", it was NOT clear what the actual question was. You "of course" is rather rude!
Oh, sorry, I didn't not intend to be rude. English is not my first language, we use to meaning of 'of course' not to be mean but just to clarify something.
blamocur
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No, it is not: [imath]\cos \frac{3\pi}{4} \neq \frac{1}{2}[/imath]
blamocur
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Sorry for the typo, I meant -1/2, i.e. [imath]\cos\frac{3\pi}{4} \neq \mathbf{-}\frac{1}{2}[/imath].
As for our disagreement, we can resolve it with a calculator, for example this one: https://www.calculator.net/scientific-calculator.html
As for our disagreement, we can resolve it with a calculator, for example this one: https://www.calculator.net/scientific-calculator.html
D
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Dr.Peterson
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What we need to figure out is, if the exercise said to solve the equation for x, what in the world could they mean by saying that the "result" (solution?) is an expression in x. This is why it makes no sense for you to say,
Others are working on helping you correct your solution; but can you please tell us how you came to the conclusion that 2sin(3x/2)*2cosx*cos(x/2) is "the real result"?
Also, you should know that there are other questions to ask about an equation besides to find x! Just stating the equation does not imply the question. We've had other cases just recently where the question was not what was assumed.
oh okay,
so the question was to determine x. As you can see in the end I got sinxcosx*(4cosx+2), which resulted in certain x values (one of which is incorrect, sorry about that), the result i am supposed to get is 2sin(3x/2)*2cosx*cos(x/2) which results in different x values. I thought it would result in same x values. So I am not particulary sure where I went wrong.
Dr.Peterson
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Who says you are supposed to get that? (And that is not what you get in the end; in the end you get the solutions!) So someone showed you work that led to different solutions by way of this expression? Then they are wrong! And I have no idea how they got that, but there are typically different ways to get the same solution, so their expression being different from yours does not in itself mean your work is wrong.
You haven't corrected your solution, but presumably you have realized that rather than [imath]x=\pm\frac{3\pi}{4}+2k\pi[/imath], you should have had [imath]x=\pm\frac{2\pi}{3}+2k\pi[/imath], right?
When you were asked if you had "plugged in your answers", that meant checking them by replacing x with those values to see if they are actual solutions. Surely you are taught to do that; that is how you can determine whether your answers, or the ones you are told are correct, are really correct. Do that. (What are the solutions "their" answer gave? I think you may find that they are actually the same solutions, just expressed differently. List them.)
Another thing you can do to check is to graph the function, and see if its x-intercepts agree with your solutions or theirs:
Does this agree with your solutions?
P.K.Tandon
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Club : sin 3x and sin x. Use the identity ( sin C + sin D ) = 2 sin (C+D)/2 * cos (C-D)/2. and simplify You will get : (sin 2x) (2cos x + 1 ) = 0 now proceed from here.