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Are all values of $x$ solutions for $e^2ln(sin(x)) = 1 - e^2ln(cos(x))$ in $mathbb R$?

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1

0

$begingroup$

Does all values of $x$ in $mathbb R$ satisfy equation: $$e^2ln(sin(x)) = 1 - e^2ln(cos(x))$$

I am asking this, because by checking WolframAlpha solution there is an answer: (all values for $x$ are solutions over reals), but we know that $ln(0)$ is undefined, same for negative numbers.

Wolfram Alpha solution

Therefore I assume in $mathbb R$, zero and negative numbers doesn't satisfy this equation.

logarithms wolfram-alpha

share|cite|improve this question

edited Mar 13 at 18:07

Maria Mazur

47.7k1260120

asked Mar 13 at 17:50

Peter ParadaPeter Parada

1086

New contributor

Peter Parada is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What exactly do you mean by $ln*sin(x)$? Does it refer to $ln(sin(x))$ or what is the meaning of this notation?
  $endgroup$
  – mrtaurho
  Mar 13 at 17:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry, my bad. Yes. Will fix equation
  $endgroup$
  – Peter Parada
  Mar 13 at 17:53
  

  
  
  
  

add a comment | 

1

0

$begingroup$

Does all values of $x$ in $mathbb R$ satisfy equation: $$e^2ln(sin(x)) = 1 - e^2ln(cos(x))$$

I am asking this, because by checking WolframAlpha solution there is an answer: (all values for $x$ are solutions over reals), but we know that $ln(0)$ is undefined, same for negative numbers.

Wolfram Alpha solution

Therefore I assume in $mathbb R$, zero and negative numbers doesn't satisfy this equation.

logarithms wolfram-alpha

share|cite|improve this question

edited Mar 13 at 18:07

Maria Mazur

47.7k1260120

asked Mar 13 at 17:50

Peter ParadaPeter Parada

1086

New contributor

Peter Parada is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What exactly do you mean by $ln*sin(x)$? Does it refer to $ln(sin(x))$ or what is the meaning of this notation?
  $endgroup$
  – mrtaurho
  Mar 13 at 17:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry, my bad. Yes. Will fix equation
  $endgroup$
  – Peter Parada
  Mar 13 at 17:53
  

  
  
  
  

add a comment | 

$begingroup$

Does all values of $x$ in $mathbb R$ satisfy equation: $$e^2ln(sin(x)) = 1 - e^2ln(cos(x))$$

I am asking this, because by checking WolframAlpha solution there is an answer: (all values for $x$ are solutions over reals), but we know that $ln(0)$ is undefined, same for negative numbers.

Wolfram Alpha solution

Therefore I assume in $mathbb R$, zero and negative numbers doesn't satisfy this equation.

logarithms wolfram-alpha

share|cite|improve this question

edited Mar 13 at 18:07

Maria Mazur

47.7k1260120

asked Mar 13 at 17:50

Peter ParadaPeter Parada

1086

New contributor

Peter Parada is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 13 at 18:07

Maria Mazur

47.7k1260120

asked Mar 13 at 17:50

Peter ParadaPeter Parada

1086

New contributor

Peter Parada is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 13 at 18:07

Maria Mazur

47.7k1260120

asked Mar 13 at 17:50

Peter ParadaPeter Parada

1086

New contributor

Peter Parada is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 13 at 18:07

Maria Mazur

47.7k1260120

edited Mar 13 at 18:07

Maria Mazur

47.7k1260120

asked Mar 13 at 17:50

Peter ParadaPeter Parada

1086

New contributor

Peter Parada is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 13 at 17:50

Peter ParadaPeter Parada

1086

2 Answers
2

active

oldest

votes

2

$begingroup$

$$1=e^2lnsin(x) +e^2ln cos(x)= e^lnsin^2(x)+e^ln cos^2(x) = $$

$$ = sin^2(x)+cos^2(x) = 1 $$

So this is true for all $x$ such that $sin x>0$ and $cos x>0$ (because then is $ln $ defined).

So $$boxedxin colorredbigcup_kin mathbbZ (0+2pi k,piover 2+2pi k)$$

share|cite|improve this answer

edited Mar 13 at 18:03

answered Mar 13 at 17:58

Maria MazurMaria Mazur

47.7k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great. I was thinking the same. Does that mean solution on Wolfram is wrong?
  $endgroup$
  – Peter Parada
  Mar 13 at 18:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) For myself I totally forgot to think about the existence of the natural logarithm as it is only assured for $sin(x),cos(x)>0$.
  $endgroup$
  – mrtaurho
  Mar 13 at 18:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterParada I'm not sure what is Wolfram doing. It draw you a whole line for $f(x) =e^ln x$ which is absurd to me.
  $endgroup$
  – Maria Mazur
  Mar 13 at 18:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I will write them for further clarification.
  $endgroup$
  – Peter Parada
  Mar 13 at 18:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterParada: I think Wolfram implicitly assumes that $x$ is in the domain; you might need to check documentation to see if this is standard.
  $endgroup$
  – Clayton
  Mar 13 at 18:08
  

  
  
  
  

add a comment | 

0

$begingroup$

Note that :
$$e^2ln sin x + e^2 ln cos x = 1 Rightarrow e^ln (sin x)^2 + e^ln (cos x)^2 = 1 Leftrightarrow sin^2x + cos^2x = 1 rightarrow texttrue ; forall x in mathbb R$$
Restrictions apply so as the initial expression holds, so that narrows down the solution set. Note the usage of $Rightarrow$ instead of an $Leftrightarrow$ at the start.

share|cite|improve this answer

edited Mar 13 at 18:04

answered Mar 13 at 17:58

RebellosRebellos

15.4k31250

$endgroup$

add a comment | 

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2

$begingroup$

$$1=e^2lnsin(x) +e^2ln cos(x)= e^lnsin^2(x)+e^ln cos^2(x) = $$

$$ = sin^2(x)+cos^2(x) = 1 $$

So this is true for all $x$ such that $sin x>0$ and $cos x>0$ (because then is $ln $ defined).

So $$boxedxin colorredbigcup_kin mathbbZ (0+2pi k,piover 2+2pi k)$$

share|cite|improve this answer

edited Mar 13 at 18:03

answered Mar 13 at 17:58

Maria MazurMaria Mazur

47.7k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great. I was thinking the same. Does that mean solution on Wolfram is wrong?
  $endgroup$
  – Peter Parada
  Mar 13 at 18:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) For myself I totally forgot to think about the existence of the natural logarithm as it is only assured for $sin(x),cos(x)>0$.
  $endgroup$
  – mrtaurho
  Mar 13 at 18:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterParada I'm not sure what is Wolfram doing. It draw you a whole line for $f(x) =e^ln x$ which is absurd to me.
  $endgroup$
  – Maria Mazur
  Mar 13 at 18:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I will write them for further clarification.
  $endgroup$
  – Peter Parada
  Mar 13 at 18:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterParada: I think Wolfram implicitly assumes that $x$ is in the domain; you might need to check documentation to see if this is standard.
  $endgroup$
  – Clayton
  Mar 13 at 18:08
  

  
  
  
  

add a comment | 

2

$begingroup$

$$1=e^2lnsin(x) +e^2ln cos(x)= e^lnsin^2(x)+e^ln cos^2(x) = $$

$$ = sin^2(x)+cos^2(x) = 1 $$

So this is true for all $x$ such that $sin x>0$ and $cos x>0$ (because then is $ln $ defined).

So $$boxedxin colorredbigcup_kin mathbbZ (0+2pi k,piover 2+2pi k)$$

share|cite|improve this answer

edited Mar 13 at 18:03

answered Mar 13 at 17:58

Maria MazurMaria Mazur

47.7k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great. I was thinking the same. Does that mean solution on Wolfram is wrong?
  $endgroup$
  – Peter Parada
  Mar 13 at 18:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) For myself I totally forgot to think about the existence of the natural logarithm as it is only assured for $sin(x),cos(x)>0$.
  $endgroup$
  – mrtaurho
  Mar 13 at 18:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterParada I'm not sure what is Wolfram doing. It draw you a whole line for $f(x) =e^ln x$ which is absurd to me.
  $endgroup$
  – Maria Mazur
  Mar 13 at 18:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I will write them for further clarification.
  $endgroup$
  – Peter Parada
  Mar 13 at 18:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterParada: I think Wolfram implicitly assumes that $x$ is in the domain; you might need to check documentation to see if this is standard.
  $endgroup$
  – Clayton
  Mar 13 at 18:08
  

  
  
  
  

add a comment | 

$begingroup$

$$1=e^2lnsin(x) +e^2ln cos(x)= e^lnsin^2(x)+e^ln cos^2(x) = $$

$$ = sin^2(x)+cos^2(x) = 1 $$

So this is true for all $x$ such that $sin x>0$ and $cos x>0$ (because then is $ln $ defined).

So $$boxedxin colorredbigcup_kin mathbbZ (0+2pi k,piover 2+2pi k)$$

share|cite|improve this answer

edited Mar 13 at 18:03

answered Mar 13 at 17:58

Maria MazurMaria Mazur

47.7k1260120

$endgroup$

share|cite|improve this answer

edited Mar 13 at 18:03

answered Mar 13 at 17:58

Maria MazurMaria Mazur

47.7k1260120

answered Mar 13 at 17:58

Maria MazurMaria Mazur

47.7k1260120

answered Mar 13 at 17:58

Maria MazurMaria Mazur

47.7k1260120

0

$begingroup$

Note that :
$$e^2ln sin x + e^2 ln cos x = 1 Rightarrow e^ln (sin x)^2 + e^ln (cos x)^2 = 1 Leftrightarrow sin^2x + cos^2x = 1 rightarrow texttrue ; forall x in mathbb R$$
Restrictions apply so as the initial expression holds, so that narrows down the solution set. Note the usage of $Rightarrow$ instead of an $Leftrightarrow$ at the start.

share|cite|improve this answer

edited Mar 13 at 18:04

answered Mar 13 at 17:58

RebellosRebellos

15.4k31250

$endgroup$

add a comment | 

0

$begingroup$

Note that :
$$e^2ln sin x + e^2 ln cos x = 1 Rightarrow e^ln (sin x)^2 + e^ln (cos x)^2 = 1 Leftrightarrow sin^2x + cos^2x = 1 rightarrow texttrue ; forall x in mathbb R$$
Restrictions apply so as the initial expression holds, so that narrows down the solution set. Note the usage of $Rightarrow$ instead of an $Leftrightarrow$ at the start.

share|cite|improve this answer

edited Mar 13 at 18:04

answered Mar 13 at 17:58

RebellosRebellos

15.4k31250

$endgroup$

add a comment | 

$begingroup$

Note that :
$$e^2ln sin x + e^2 ln cos x = 1 Rightarrow e^ln (sin x)^2 + e^ln (cos x)^2 = 1 Leftrightarrow sin^2x + cos^2x = 1 rightarrow texttrue ; forall x in mathbb R$$
Restrictions apply so as the initial expression holds, so that narrows down the solution set. Note the usage of $Rightarrow$ instead of an $Leftrightarrow$ at the start.

share|cite|improve this answer

edited Mar 13 at 18:04

answered Mar 13 at 17:58

RebellosRebellos

15.4k31250

$endgroup$

share|cite|improve this answer

edited Mar 13 at 18:04

answered Mar 13 at 17:58

RebellosRebellos

15.4k31250

answered Mar 13 at 17:58

RebellosRebellos

15.4k31250

answered Mar 13 at 17:58

RebellosRebellos

15.4k31250