Finding the solutions of equation over interval

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rachelmaddie

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Hi. I need help with this problem.
6E869299-4EA5-4DBE-8B63-CA391390E2A9.png

I got x = 4/3, 8/3, 4, 16/3

But I need help with showing my work because I used mathway for help with this.
 
Can't you just simplify the LHS using the fact that A + A = 2A? Then divide by 2, and so on. It seems trivial to me -- in fact, too trivial to be a real problem!

And the solution of that problem is not what you say. Was there a typo somewhere?
 
Dr.Peterson said:
Can't you just simplify the LHS using the fact that A + A = 2A? Then divide by 2, and so on. It seems trivial to me -- in fact, too trivial to be a real problem!

And the solution of that problem is not what you say. Was there a typo somewhere?
Click to expand...
-cosx - cosx = - 2
-2cosx = -2
cosx = 1
x = 0
Like this?
 
Why did you not do that in the first place? Those two daunting expressions on the LHS look awfully compelling.
 
rachelmaddie said:
the general solution is 360 × n
Click to expand...
Which of course means that the only solution in the stated interval is 0.

It's worth noting that in this context, the plural is hypothetical! It is possible for "all solutions" to be one solution, or even none.

By the way, I solved it a little differently: [MATH]\sin(3\pi/2 + x) = -1[/MATH], so [MATH]3\pi/2 + x = 3\pi/2 + 2n\pi[/MATH], and [MATH]x = 2n\pi[/MATH]. (You gave this in terms of degrees, which is of course wrong, as the equation is written with radians.)
 
Dr.Peterson said:
Which of course means that the only solution in the stated interval is 0.

It's worth noting that in this context, the plural is hypothetical! It is possible for "all solutions" to be one solution, or even none.

By the way, I solved it a little differently: [MATH]\sin(3\pi/2 + x) = -1[/MATH], so [MATH]3\pi/2 + x = 3\pi/2 + 2n\pi[/MATH], and [MATH]x = 2n\pi[/MATH]. (You gave this in terms of degrees, which is of course wrong, as the equation is written with radians.)
Click to expand...
Can you please check this?

-cosx - cosx = - 2
-2cosx = -2
cosx = 1
x = 0
sin(3pi/2 + x) = -1, so 3pi/2 + x = 3pi/2 + 2npi, and x = 2npi
The only solution in the stated interval is 0. It is possible for "all solutions" to be one solution, or even none.
 
You don't need to show both your solution and mine; either alone is enough!

To show that your solution, x = 0, is the only one in the given interval, you just have to know that the cosine is only 1 at one place in a cycle, namely the beginning (which is the same as the end, the start of the next cycle).

Alternatively, you could do as I did and state explicitly the general solution and then point out that only one solution is in the first cycle:

-cosx - cosx = - 2​
-2cosx = -2​
cosx = 1​
x = 0 + 2n pi (for any integer n) -- general solution​
x = 0 -- only solution in [0, 2 pi)​
 
Dr.Peterson said:
You don't need to show both your solution and mine; either alone is enough!

To show that your solution, x = 0, is the only one in the given interval, you just have to know that the cosine is only 1 at one place in a cycle, namely the beginning (which is the same as the end, the start of the next cycle).

Alternatively, you could do as I did and state explicitly the general solution and then point out that only one solution is in the first cycle:

-cosx - cosx = - 2​
-2cosx = -2​
cosx = 1​
x = 0 + 2n pi (for any integer n) -- general solution​
x = 0 -- only solution in [0, 2 pi)​
Click to expand...
Like this?

-cosx - cosx = - 2
-2cosx = -2
cosx = 1
sin(3pi/2 + x) = -1, so 3pi/2 + x = 3pi/2 + 2npi, and x = 2npi
x = 0 + 2n pi (for any integer n) -- general solution
x = 0 -- only solution in [0, 2 pi)
 
rachelmaddie said:
Like this?

-cosx - cosx = - 2
-2cosx = -2
cosx = 1
sin(3pi/2 + x) = -1, so 3pi/2 + x = 3pi/2 + 2npi, and x = 2npi
x = 0 + 2n pi (for any integer n) -- general solution
x = 0 -- only solution in [0, 2 pi)
Click to expand...
No. Drop the line about the sine, which doesn't belong there at all. That is my alternate solution, not part of yours.

Please think about how the lines you write fit together, rather than blindly copying things as given to you and sticking them together. The goal here is understanding, and you seem to be avoiding that.
 
Dr.Peterson said:
No. Drop the line about the sine, which doesn't belong there at all. That is my alternate solution, not part of yours.

Please think about how the lines you write fit together, rather than blindly copying things as given to you and sticking them together. The goal here is understanding, and you seem to be avoiding that.
Click to expand...
-cosx - cosx = - 2
-2cosx = -2
cosx = 1
x = 0 + 2n pi (for any integer n) -- general solution
x = 0 -- only solution in [0, 2 pi)
The only solution in the stated interval is 0. It is possible for "all solutions" to be one solution, or even none.

Should I include this statement?
 
The last sentence was just my general comment, and is not part of the answer to the question.

But you can make your own decisions about how to say things. It is not my intention to dictate to you the one best way to answer a question.
 
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