Why can't I square it?

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Loki123

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I tried to do this problem two ways. First is incorrect, second is correct. I am having problems with understanding why I can't square it. IMG_20220102_182408.jpg
 
Loki123 said:
I tried to do this problem two ways. First is incorrect, second is correct. I am having problems with understanding why I can't square it. View attachment 30437
Click to expand...
I like your first way more; it's a lot easier. But you haven't finished it!

When you square an equation, you can introduce extraneous solutions. (Surely you've learned that!) In particular, your solutions will include solutions of [imath]\sin x + \cos x = 1[/imath] as well as of [imath]\sin x + \cos x = -1[/imath]. So you need to check each case and eliminate extraneous solutions. For example, if [imath]\sin x=0[/imath], then in order to satisfy the original equation, you must have [imath]\cos x = -1[/imath]. When is that true?
 
Another approach you can take if you notice 2sin(x)cos(x)=sin(2x)=0, then 2x=arcsin(0). If you want to take this route, continue the problem and post back.
 
Dr.Peterson said:
I like your first way more; it's a lot easier. But you haven't finished it!

When you square an equation, you can introduce extraneous solutions. (Surely you've learned that!) In particular, your solutions will include solutions of [imath]\sin x + \cos x = 1[/imath] as well as of [imath]\sin x + \cos x = -1[/imath]. So you need to check each case and eliminate extraneous solutions. For example, if [imath]\sin x=0[/imath], then in order to satisfy the original equation, you must have [imath]\cos x = -1[/imath]. When is that true?
Click to expand...
IMG_20220102_201613.jpg
What now?
 
What did the problem ask you to FIND?

What is the question you needed to answer?
 
Dr.Peterson said:
Now, observe that this is equivalent to the answer you got the other way!

Or. compare both to the solutions given by a graph:
View attachment 30442
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I am a little confused. So I basically got four x. Two are conditions for sinx=0 or cosx=0. Don't I need to find where they overlap or something? Do I just leave the four of them?
 
Loki123 said:
I am a little confused. So I basically got four x. Two are conditions for sinx=0 or cosx=0. Don't I need to find where they overlap or something? Do I just leave the four of them?
Click to expand...
Huh? You have infinitely many solutions, defined by two expressions. What do you mean by "four x"? How is an "x" a "condition"? What do you mean by "overlap"? Please explain your thinking more fully.
 
Dr.Peterson said:
Huh? You have infinitely many solutions, defined by two expressions. What do you mean by "four x"? How is an "x" a "condition"? What do you mean by "overlap"? Please explain your thinking more fully.
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Sure, so yes there are infinitely many solutions. I mean that in my last picture i wrote what x equals four times. (I outlined them) First x is what x needs to be for sinx=0, second x is what x needs to be for cosx=0, third x is what cosx needs to be if sinx=0 so we can have -1 when we add them together, four x is what sinx needs to be if cosx=0 so we can have -1 when we add them together.
I am confused what I can write as the final solution. See if I have one x for sinx=0 and one x for cosx=-1, do i not need to find a solution that works for both of them?
 
Loki123 said:
I am a little confused. So I basically got four x. Two are conditions for sinx=0 or cosx=0. Don't I need to find where they overlap or something? Do I just leave the four of them?
Click to expand...
If you look back at your original solutions in the OP
[imath]x= \pi k[/imath] and [imath]x=\frac{\pi}{2} + \pi k[/imath]. Notice that your solutions lie on the x and y axis, meaning the solutions occur at every [imath]\frac{\pi k}{2}[/imath]
 
BigBeachBananas said:
If you look back at your original solutions in the OP
[imath]x= \pi k[/imath] and [imath]x=\frac{\pi}{2} + \pi k[/imath]. Notice that your solutions lie on the x and y axis, meaning the solutions occur at every [imath]\frac{\pi k}{2}[/imath]
Click to expand...
those are incorrect. solutions from second method are correct
 
Loki123 said:
Sure, so yes there are infinitely many solutions. I mean that in my last picture i wrote what x equals four times. (I outlined them) First x is what x needs to be for sinx=0, second x is what x needs to be for cosx=0, third x is what cosx needs to be if sinx=0 so we can have -1 when we add them together, four x is what sinx needs to be if cosx=0 so we can have -1 when we add them together.
I am confused what I can write as the final solution. See if I have one x for sinx=0 and one x for cosx=-1, do i not need to find a solution that works for both of them?
Click to expand...
Here is what you showed:
1641155812667.png

The top pair of boxed solutions are the initial solutions that include extraneous solutions.

The bottom pair are the non-extraneous solutions within the initial solutions. The extraneous solutions have been removed, by ensuring that the original equation is satisfied, so you can ignore the top pair. You have only two sets of solutions here.

Since EITHER [imath]\sin x = 0[/imath] OR [imath]\cos x = 0[/imath], the full solution is the union of these, not the intersection, if the latter is what you mean by "works for both of them". The solution is that EITHER [imath]x=\pi+2k\pi[/imath] OR [imath]x=\frac{3\pi}{2}+2k\pi[/imath]. And this agrees with your second method:
1641156131991.png
There, too, the solution is that x can be any of the solutions described (their union).
 
Dr.Peterson said:
Here is what you showed:
View attachment 30446

The top pair of boxed solutions are the initial solutions that include extraneous solutions.

The bottom pair are the non-extraneous solutions within the initial solutions. The extraneous solutions have been removed, by ensuring that the original equation is satisfied, so you can ignore the top pair. You have only two sets of solutions here.

Since EITHER [imath]\sin x = 0[/imath] OR [imath]\cos x = 0[/imath], the full solution is the union of these, not the intersection, if the latter is what you mean by "works for both of them". The solution is that EITHER [imath]x=\pi+2k\pi[/imath] OR [imath]x=\frac{3\pi}{2}+2k\pi[/imath]. And this agrees with your second method:
View attachment 30447
There, too, the solution is that x can be any of the solutions described (their union).
Click to expand...
Okay! Thank you! I understand now.
 
Dr.Peterson said:
Here is what you showed:
View attachment 30446

The top pair of boxed solutions are the initial solutions that include extraneous solutions.

The bottom pair are the non-extraneous solutions within the initial solutions. The extraneous solutions have been removed, by ensuring that the original equation is satisfied, so you can ignore the top pair. You have only two sets of solutions here.

Since EITHER [imath]\sin x = 0[/imath] OR [imath]\cos x = 0[/imath], the full solution is the union of these, not the intersection, if the latter is what you mean by "works for both of them". The solution is that EITHER [imath]x=\pi+2k\pi[/imath] OR [imath]x=\frac{3\pi}{2}+2k\pi[/imath]. And this agrees with your second method:
View attachment 30447
There, too, the solution is that x can be any of the solutions described (their union).
Click to expand...
one side question, say I solve some equation through sinx, but the result from the book solved it through cosx. The x will obviously be written differently, how do I check if it's the same?
 
Loki123 said:
one side question, say I solve some equation through sinx, but the result from the book solved it through cosx. The x will obviously be written differently, how do I check if it's the same?
Click to expand...
For something like the current example, just check whether each solution written one way is in the set stated the other way. (Or, if you just want a quick check, graph both answers and see if they agree, as I showed in #6.)

It might be a good habit, however, to write all solutions in a parallel way; in this example, your
1641219294055.png​
was not written in a simplified form, and when simplified uses a negative angle, to which you can add [imath]2\pi[/imath] to get it into the positive range you used in your other equivalent answer. This would make the final comparison easy.

In other cases, your answer to a problem may actually be written in terms of different trig functions (rather than being just numbers, as here). In that case, treat the claim that they are equal as an identity, and prove it! (Or, again, graph both.)

This is a very common issue in trig, because everything can be stated in many ways.
 
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