Find number of possible values satisfying $cos left(pisqrt{x-4}right)cos left(pisqrt{x}right)=1$ [closed]

-1

$begingroup$

Find number of possible values satisfying the equation:
$$cos left(pisqrt{x-4}right)cos left(pisqrt{x}right)=1$$

We are required to find the number of possible solutions for $xinBbb R$.

I can't understand how to approach this.

trigonometry

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edited Jan 21 at 11:35

Blue

48.8k870156

asked Jan 21 at 11:23

Sameer ThakurSameer Thakur

305

$endgroup$

closed as off-topic by Eevee Trainer, Jose Arnaldo Bebita Dris, Adrian Keister, Abcd, Shaun Jan 21 at 15:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Jose Arnaldo Bebita Dris, Adrian Keister, Abcd, Shaun

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might want to learn the difference between "can't" and "don't" :)
  $endgroup$
  – Shaun
  Jan 21 at 15:31
  

  
  
  
  

add a comment | 

-1

$begingroup$

Find number of possible values satisfying the equation:
$$cos left(pisqrt{x-4}right)cos left(pisqrt{x}right)=1$$

We are required to find the number of possible solutions for $xinBbb R$.

I can't understand how to approach this.

trigonometry

share|cite|improve this question

edited Jan 21 at 11:35

Blue

48.8k870156

asked Jan 21 at 11:23

Sameer ThakurSameer Thakur

305

$endgroup$

closed as off-topic by Eevee Trainer, Jose Arnaldo Bebita Dris, Adrian Keister, Abcd, Shaun Jan 21 at 15:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Jose Arnaldo Bebita Dris, Adrian Keister, Abcd, Shaun

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might want to learn the difference between "can't" and "don't" :)
  $endgroup$
  – Shaun
  Jan 21 at 15:31
  

  
  
  
  

add a comment | 

-1

-1

-1

$begingroup$

Find number of possible values satisfying the equation:
$$cos left(pisqrt{x-4}right)cos left(pisqrt{x}right)=1$$

We are required to find the number of possible solutions for $xinBbb R$.

I can't understand how to approach this.

trigonometry

share|cite|improve this question

edited Jan 21 at 11:35

Blue

48.8k870156

asked Jan 21 at 11:23

Sameer ThakurSameer Thakur

305

$endgroup$

Find number of possible values satisfying the equation:
$$cos left(pisqrt{x-4}right)cos left(pisqrt{x}right)=1$$

We are required to find the number of possible solutions for $xinBbb R$.

I can't understand how to approach this.

trigonometry

trigonometry

share|cite|improve this question

edited Jan 21 at 11:35

Blue

48.8k870156

asked Jan 21 at 11:23

Sameer ThakurSameer Thakur

305

share|cite|improve this question

edited Jan 21 at 11:35

Blue

48.8k870156

asked Jan 21 at 11:23

Sameer ThakurSameer Thakur

305

share|cite|improve this question

share|cite|improve this question

edited Jan 21 at 11:35

Blue

48.8k870156

edited Jan 21 at 11:35

Blue

48.8k870156

edited Jan 21 at 11:35

Blue

48.8k870156

48.8k870156

asked Jan 21 at 11:23

Sameer ThakurSameer Thakur

305

asked Jan 21 at 11:23

Sameer ThakurSameer Thakur

305

asked Jan 21 at 11:23

Sameer ThakurSameer Thakur

305

305

closed as off-topic by Eevee Trainer, Jose Arnaldo Bebita Dris, Adrian Keister, Abcd, Shaun Jan 21 at 15:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Jose Arnaldo Bebita Dris, Adrian Keister, Abcd, Shaun

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Eevee Trainer, Jose Arnaldo Bebita Dris, Adrian Keister, Abcd, Shaun Jan 21 at 15:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Jose Arnaldo Bebita Dris, Adrian Keister, Abcd, Shaun

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might want to learn the difference between "can't" and "don't" :)
  $endgroup$
  – Shaun
  Jan 21 at 15:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might want to learn the difference between "can't" and "don't" :)
  $endgroup$
  – Shaun
  Jan 21 at 15:31
  

  
  
  
  

$begingroup$
You might want to learn the difference between "can't" and "don't" :)
$endgroup$
– Shaun
Jan 21 at 15:31

$begingroup$
You might want to learn the difference between "can't" and "don't" :)
$endgroup$
– Shaun
Jan 21 at 15:31

add a comment | 

1 Answer
1

active

oldest

votes

2

$begingroup$

Hint:

Since $cos theta in [-1,1]$, and we're given the equation

$$cos(pi sqrt{x - 4}) cos(pi sqrt x) = 1$$

then either

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = 1$$

or

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = -1$$

Don't forget to invoke the periodicity of cosine as well.

share|cite|improve this answer

answered Jan 21 at 11:32

Eevee TrainerEevee Trainer

7,42821338

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do not forget that $cos(it)=cosh(t)$ which makes the problem interesting.
  $endgroup$
  – Claude Leibovici
  Jan 21 at 11:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you conclude that these two are the only possible cases?
  $endgroup$
  – Sameer Thakur
  Jan 21 at 13:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It partly is from we know cosine cannot be outside $[-1,1]$. If two numbers $a,b$ satisfy $$a cdot b = 1$$ then it follows $$a = frac 1 b ;;; b = frac 1 a$$ Obviously, if $b<1$ (left) or $a<1$ (right), then the right side is greater than $1$. Since both $a,b$ in this case are cosine functions, it follows we cannot have either be less than one. Thus, their product is $1$, as given, which would only be achieved if both were $1$ or $-1$.
  $endgroup$
  – Eevee Trainer
  Jan 21 at 13:34
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

Hint:

Since $cos theta in [-1,1]$, and we're given the equation

$$cos(pi sqrt{x - 4}) cos(pi sqrt x) = 1$$

then either

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = 1$$

or

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = -1$$

Don't forget to invoke the periodicity of cosine as well.

share|cite|improve this answer

answered Jan 21 at 11:32

Eevee TrainerEevee Trainer

7,42821338

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do not forget that $cos(it)=cosh(t)$ which makes the problem interesting.
  $endgroup$
  – Claude Leibovici
  Jan 21 at 11:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you conclude that these two are the only possible cases?
  $endgroup$
  – Sameer Thakur
  Jan 21 at 13:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It partly is from we know cosine cannot be outside $[-1,1]$. If two numbers $a,b$ satisfy $$a cdot b = 1$$ then it follows $$a = frac 1 b ;;; b = frac 1 a$$ Obviously, if $b<1$ (left) or $a<1$ (right), then the right side is greater than $1$. Since both $a,b$ in this case are cosine functions, it follows we cannot have either be less than one. Thus, their product is $1$, as given, which would only be achieved if both were $1$ or $-1$.
  $endgroup$
  – Eevee Trainer
  Jan 21 at 13:34
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint:

Since $cos theta in [-1,1]$, and we're given the equation

$$cos(pi sqrt{x - 4}) cos(pi sqrt x) = 1$$

then either

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = 1$$

or

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = -1$$

Don't forget to invoke the periodicity of cosine as well.

share|cite|improve this answer

answered Jan 21 at 11:32

Eevee TrainerEevee Trainer

7,42821338

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do not forget that $cos(it)=cosh(t)$ which makes the problem interesting.
  $endgroup$
  – Claude Leibovici
  Jan 21 at 11:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you conclude that these two are the only possible cases?
  $endgroup$
  – Sameer Thakur
  Jan 21 at 13:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It partly is from we know cosine cannot be outside $[-1,1]$. If two numbers $a,b$ satisfy $$a cdot b = 1$$ then it follows $$a = frac 1 b ;;; b = frac 1 a$$ Obviously, if $b<1$ (left) or $a<1$ (right), then the right side is greater than $1$. Since both $a,b$ in this case are cosine functions, it follows we cannot have either be less than one. Thus, their product is $1$, as given, which would only be achieved if both were $1$ or $-1$.
  $endgroup$
  – Eevee Trainer
  Jan 21 at 13:34
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

Hint:

Since $cos theta in [-1,1]$, and we're given the equation

$$cos(pi sqrt{x - 4}) cos(pi sqrt x) = 1$$

then either

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = 1$$

or

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = -1$$

Don't forget to invoke the periodicity of cosine as well.

share|cite|improve this answer

answered Jan 21 at 11:32

Eevee TrainerEevee Trainer

7,42821338

$endgroup$

Hint:

Since $cos theta in [-1,1]$, and we're given the equation

$$cos(pi sqrt{x - 4}) cos(pi sqrt x) = 1$$

then either

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = 1$$

or

$$cos(pi sqrt{x - 4}) = cos(pi sqrt x) = -1$$

Don't forget to invoke the periodicity of cosine as well.

share|cite|improve this answer

answered Jan 21 at 11:32

Eevee TrainerEevee Trainer

7,42821338

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 at 11:32

Eevee TrainerEevee Trainer

7,42821338

answered Jan 21 at 11:32

Eevee TrainerEevee Trainer

7,42821338

answered Jan 21 at 11:32

Eevee TrainerEevee Trainer

7,42821338

7,42821338

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do not forget that $cos(it)=cosh(t)$ which makes the problem interesting.
  $endgroup$
  – Claude Leibovici
  Jan 21 at 11:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you conclude that these two are the only possible cases?
  $endgroup$
  – Sameer Thakur
  Jan 21 at 13:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It partly is from we know cosine cannot be outside $[-1,1]$. If two numbers $a,b$ satisfy $$a cdot b = 1$$ then it follows $$a = frac 1 b ;;; b = frac 1 a$$ Obviously, if $b<1$ (left) or $a<1$ (right), then the right side is greater than $1$. Since both $a,b$ in this case are cosine functions, it follows we cannot have either be less than one. Thus, their product is $1$, as given, which would only be achieved if both were $1$ or $-1$.
  $endgroup$
  – Eevee Trainer
  Jan 21 at 13:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do not forget that $cos(it)=cosh(t)$ which makes the problem interesting.
  $endgroup$
  – Claude Leibovici
  Jan 21 at 11:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you conclude that these two are the only possible cases?
  $endgroup$
  – Sameer Thakur
  Jan 21 at 13:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It partly is from we know cosine cannot be outside $[-1,1]$. If two numbers $a,b$ satisfy $$a cdot b = 1$$ then it follows $$a = frac 1 b ;;; b = frac 1 a$$ Obviously, if $b<1$ (left) or $a<1$ (right), then the right side is greater than $1$. Since both $a,b$ in this case are cosine functions, it follows we cannot have either be less than one. Thus, their product is $1$, as given, which would only be achieved if both were $1$ or $-1$.
  $endgroup$
  – Eevee Trainer
  Jan 21 at 13:34
  

  
  
  
  

1

1

$begingroup$
Do not forget that $cos(it)=cosh(t)$ which makes the problem interesting.
$endgroup$
– Claude Leibovici
Jan 21 at 11:37

$begingroup$
Do not forget that $cos(it)=cosh(t)$ which makes the problem interesting.
$endgroup$
– Claude Leibovici
Jan 21 at 11:37

$begingroup$
How did you conclude that these two are the only possible cases?
$endgroup$
– Sameer Thakur
Jan 21 at 13:16

$begingroup$
How did you conclude that these two are the only possible cases?
$endgroup$
– Sameer Thakur
Jan 21 at 13:16

$begingroup$
It partly is from we know cosine cannot be outside $[-1,1]$. If two numbers $a,b$ satisfy $$a cdot b = 1$$ then it follows $$a = frac 1 b ;;; b = frac 1 a$$ Obviously, if $b<1$ (left) or $a<1$ (right), then the right side is greater than $1$. Since both $a,b$ in this case are cosine functions, it follows we cannot have either be less than one. Thus, their product is $1$, as given, which would only be achieved if both were $1$ or $-1$.
$endgroup$
– Eevee Trainer
Jan 21 at 13:34

$begingroup$
It partly is from we know cosine cannot be outside $[-1,1]$. If two numbers $a,b$ satisfy $$a cdot b = 1$$ then it follows $$a = frac 1 b ;;; b = frac 1 a$$ Obviously, if $b<1$ (left) or $a<1$ (right), then the right side is greater than $1$. Since both $a,b$ in this case are cosine functions, it follows we cannot have either be less than one. Thus, their product is $1$, as given, which would only be achieved if both were $1$ or $-1$.
$endgroup$
– Eevee Trainer
Jan 21 at 13:34

add a comment | 

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