Different solutions to one trig equation?

skeptipus

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Dec 12, 2018
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Hello

Please look at my attempted solutions.


  1. For Question 2, are both of my solutions correct?
  2. For Question 3, what am I doing wrong? The official solution has two sets of answers. I have only one.
  3. For Question 4, my answer is very different from the actual answer. The difference arises only due to a sign change. What am I doing wrong?

Thank you very much for your time.
 

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Last edited:
Please look at my attempted solutions, and the official answers to the questions.

  1. For Question 2, are both of my solutions correct?
  2. For Question 3, what am I doing wrong? The official solution has two sets of answers. I have only one.
  3. For Question 4, my answer is very different from the actual answer. The difference arises only due to a sign change. What am I doing wrong?
Because of the layout of your post it is hard to follow your questions.
That said, in #3 you divided by the \(\displaystyle \sin(x)\) which eliminated a complete set of solutions.
That should show why it is not advisable to do that. That is the solutions by factoring is the best way to approach these.
 
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Because of the layout of your post it is hard to follow your questions.
That said, in #3 you divided by the \(\displaystyle \sin(x)\) which eliminated a complete set of solutions.
That should show why it is not advisable to do that. That is the solutions by factoring is the best way to approach these.

Thank you for your answer! I see my error in Q3 now.

I've edited my post. Initially, I was attempting to keep the attachments' quality from getting downgraded...
 
  1. For Question 4, my answer is very different from the actual answer. The difference arises only due to a sign change. What am I doing wrong?

When AB = 2, you can't conclude that either A = 2 or B = 2! You can only do that when the product is 0.

In fact, if AB = 2, all you can conclude is that neither A nor B can be 0. They could be 1 and 2, or 1/2 and 4, or 100 and 1/50, or whatever. This is why we put a zero on the right before factoring -- "standard form". Which is, of course, what they did.
 
Hello

Please look at my attempted solutions.


  1. For Question 2, are both of my solutions correct?
  2. For Question 3, what am I doing wrong? The official solution has two sets of answers. I have only one.
  3. For Question 4, my answer is very different from the actual answer. The difference arises only due to a sign change. What am I doing wrong?

Thank you very much for your time.
RE #4. For a product of two factors to equals 2, it does not imply that one of the factors is 2. After all, sqrt(2)*sqrt(2) = 2, 9*(2/9) = 2, (13/2)* (4/13) = 2. Most importantly, if one of the factors is 2, does not mean the product is 2.

This only works if the product equals 0 (the technical term is that the real numbers form an integral domain). That is anything real number times 0 = 0 and 0 times any real number = 0. Again note that it is NOT true that 2 times any number is 2. This is why you need to set the equation to 0 and then factor. ONLY THEN can you solve for each factor equaling 0
 
When AB = 2, you can't conclude that either A = 2 or B = 2! You can only do that when the product is 0.

In fact, if AB = 2, all you can conclude is that neither A nor B can be 0. They could be 1 and 2, or 1/2 and 4, or 100 and 1/50, or whatever. This is why we put a zero on the right before factoring -- "standard form". Which is, of course, what they did.

Got it! Thank you very much!!
 
RE #4. For a product of two factors to equals 2, it does not imply that one of the factors is 2. After all, sqrt(2)*sqrt(2) = 2, 9*(2/9) = 2, (13/2)* (4/13) = 2. Most importantly, if one of the factors is 2, does not mean the product is 2.

This only works if the product equals 0 (the technical term is that the real numbers form an integral domain). That is anything real number times 0 = 0 and 0 times any real number = 0. Again note that it is NOT true that 2 times any number is 2. This is why you need to set the equation to 0 and then factor. ONLY THEN can you solve for each factor equaling 0

I understand perfectly now.
Thank you very much for your time!!
 
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