$sin(x)+cos(x)+sin(2x)>1$ [closed]

-2

$begingroup$

I am having trouble solving this problem

http://tinypic.com/view.php?pic=vsl5s5&s=9

In our books, we have solutions in the back and solutions for $x$ are different from mine.

I get solutions that

$x>kpi$ and $x<(pi/2)+kpi$

and they get

$x>2kpi$ and $x<(pi/2)+2kpi$.

Thanks in advance

trigonometry

share|cite|improve this question

edited Jan 20 at 11:35

user144410

1,0432719

asked Jan 20 at 11:09

Rene IvetićRene Ivetić

1

$endgroup$

closed as off-topic by Nosrati, Cesareo, José Carlos Santos, max_zorn, Leucippus Jan 21 at 3:57

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." – Nosrati, Cesareo, José Carlos Santos, max_zorn, Leucippus

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MathSE. Please type your work rather than posting an image since links to an image may be deleted (and your image is sideways, making it difficult to read). This tutorial explains how to typeset mathematics on this site.
  $endgroup$
  – N. F. Taussig
  Jan 20 at 11:13
  

  
  
  
  

add a comment | 

-2

$begingroup$

I am having trouble solving this problem

http://tinypic.com/view.php?pic=vsl5s5&s=9

In our books, we have solutions in the back and solutions for $x$ are different from mine.

I get solutions that

$x>kpi$ and $x<(pi/2)+kpi$

and they get

$x>2kpi$ and $x<(pi/2)+2kpi$.

Thanks in advance

trigonometry

share|cite|improve this question

edited Jan 20 at 11:35

user144410

1,0432719

asked Jan 20 at 11:09

Rene IvetićRene Ivetić

1

$endgroup$

closed as off-topic by Nosrati, Cesareo, José Carlos Santos, max_zorn, Leucippus Jan 21 at 3:57

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." – Nosrati, Cesareo, José Carlos Santos, max_zorn, Leucippus

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MathSE. Please type your work rather than posting an image since links to an image may be deleted (and your image is sideways, making it difficult to read). This tutorial explains how to typeset mathematics on this site.
  $endgroup$
  – N. F. Taussig
  Jan 20 at 11:13
  

  
  
  
  

add a comment | 

-2

-2

-2

$begingroup$

I am having trouble solving this problem

http://tinypic.com/view.php?pic=vsl5s5&s=9

In our books, we have solutions in the back and solutions for $x$ are different from mine.

I get solutions that

$x>kpi$ and $x<(pi/2)+kpi$

and they get

$x>2kpi$ and $x<(pi/2)+2kpi$.

Thanks in advance

trigonometry

share|cite|improve this question

edited Jan 20 at 11:35

user144410

1,0432719

asked Jan 20 at 11:09

Rene IvetićRene Ivetić

1

$endgroup$

I am having trouble solving this problem

http://tinypic.com/view.php?pic=vsl5s5&s=9

In our books, we have solutions in the back and solutions for $x$ are different from mine.

I get solutions that

$x>kpi$ and $x<(pi/2)+kpi$

and they get

$x>2kpi$ and $x<(pi/2)+2kpi$.

Thanks in advance

trigonometry

trigonometry

share|cite|improve this question

edited Jan 20 at 11:35

user144410

1,0432719

asked Jan 20 at 11:09

Rene IvetićRene Ivetić

1

share|cite|improve this question

edited Jan 20 at 11:35

user144410

1,0432719

asked Jan 20 at 11:09

Rene IvetićRene Ivetić

1

share|cite|improve this question

share|cite|improve this question

edited Jan 20 at 11:35

user144410

1,0432719

edited Jan 20 at 11:35

user144410

1,0432719

edited Jan 20 at 11:35

user144410

1,0432719

1,0432719

asked Jan 20 at 11:09

Rene IvetićRene Ivetić

1

asked Jan 20 at 11:09

Rene IvetićRene Ivetić

1

asked Jan 20 at 11:09

Rene IvetićRene Ivetić

1

1

closed as off-topic by Nosrati, Cesareo, José Carlos Santos, max_zorn, Leucippus Jan 21 at 3:57

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." – Nosrati, Cesareo, José Carlos Santos, max_zorn, Leucippus

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

closed as off-topic by Nosrati, Cesareo, José Carlos Santos, max_zorn, Leucippus Jan 21 at 3:57

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." – Nosrati, Cesareo, José Carlos Santos, max_zorn, Leucippus

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MathSE. Please type your work rather than posting an image since links to an image may be deleted (and your image is sideways, making it difficult to read). This tutorial explains how to typeset mathematics on this site.
  $endgroup$
  – N. F. Taussig
  Jan 20 at 11:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MathSE. Please type your work rather than posting an image since links to an image may be deleted (and your image is sideways, making it difficult to read). This tutorial explains how to typeset mathematics on this site.
  $endgroup$
  – N. F. Taussig
  Jan 20 at 11:13
  

  
  
  
  

1

1

$begingroup$
Welcome to MathSE. Please type your work rather than posting an image since links to an image may be deleted (and your image is sideways, making it difficult to read). This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 20 at 11:13

$begingroup$
Welcome to MathSE. Please type your work rather than posting an image since links to an image may be deleted (and your image is sideways, making it difficult to read). This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 20 at 11:13

add a comment | 

2 Answers
2

active

oldest

votes

1

$begingroup$

We can make a substitution to solve this problem. By letting $t=sin{(x)}+cos{(x)}$ the problem simplifies to: $$t+t^2-1gt1$$$$t^2+t-2 gt 0$$$$(t+2)(t-1)gt0$$
For which the solutions are $t gt 1$ and $t lt -2$. So we then have for $t gt 1$:$$sin{(x)}+cos{(x)} gt 1$$$$sqrt{2} sin{(x)}gt1$$$$sin{(x)} gt frac{1}{sqrt{2}}$$$$2πnlt x lt 2πn+frac{π}{2}, n in mathbb{Z}$$
Or for $t lt -2$:$$sin{(x)}+cos{(x)} lt -2$$$$sqrt{2} sin{(x)}lt-2$$$$sin{(x)} lt -sqrt{2}$$
For which there are no solutions where $x in mathbb{R}$.

share|cite|improve this answer

answered Jan 20 at 11:25

Peter ForemanPeter Foreman

2,7271214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  begin{align*} sin x + cos x & = sqrt{2}left(frac{1}{sqrt{2}}sin x + frac{1}{sqrt{2}}cos xright)\ & = sqrt{2}left[sin xcosleft(frac{pi}{4}right) + cos xsinleft(frac{pi}{4}right)right]\ & = sqrt{2}sinleft(x + frac{pi}{4}right) end{align*}
  $endgroup$
  – N. F. Taussig
  Jan 20 at 12:49
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Substitute $$t=sin(x)+cos(x)$$ and square it and note that $$sin(2x)=2sin(x)cos(x)$$

share|cite|improve this answer

answered Jan 20 at 11:21

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

We can make a substitution to solve this problem. By letting $t=sin{(x)}+cos{(x)}$ the problem simplifies to: $$t+t^2-1gt1$$$$t^2+t-2 gt 0$$$$(t+2)(t-1)gt0$$
For which the solutions are $t gt 1$ and $t lt -2$. So we then have for $t gt 1$:$$sin{(x)}+cos{(x)} gt 1$$$$sqrt{2} sin{(x)}gt1$$$$sin{(x)} gt frac{1}{sqrt{2}}$$$$2πnlt x lt 2πn+frac{π}{2}, n in mathbb{Z}$$
Or for $t lt -2$:$$sin{(x)}+cos{(x)} lt -2$$$$sqrt{2} sin{(x)}lt-2$$$$sin{(x)} lt -sqrt{2}$$
For which there are no solutions where $x in mathbb{R}$.

share|cite|improve this answer

answered Jan 20 at 11:25

Peter ForemanPeter Foreman

2,7271214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  begin{align*} sin x + cos x & = sqrt{2}left(frac{1}{sqrt{2}}sin x + frac{1}{sqrt{2}}cos xright)\ & = sqrt{2}left[sin xcosleft(frac{pi}{4}right) + cos xsinleft(frac{pi}{4}right)right]\ & = sqrt{2}sinleft(x + frac{pi}{4}right) end{align*}
  $endgroup$
  – N. F. Taussig
  Jan 20 at 12:49
  

  
  
  
  

add a comment | 

1

$begingroup$

We can make a substitution to solve this problem. By letting $t=sin{(x)}+cos{(x)}$ the problem simplifies to: $$t+t^2-1gt1$$$$t^2+t-2 gt 0$$$$(t+2)(t-1)gt0$$
For which the solutions are $t gt 1$ and $t lt -2$. So we then have for $t gt 1$:$$sin{(x)}+cos{(x)} gt 1$$$$sqrt{2} sin{(x)}gt1$$$$sin{(x)} gt frac{1}{sqrt{2}}$$$$2πnlt x lt 2πn+frac{π}{2}, n in mathbb{Z}$$
Or for $t lt -2$:$$sin{(x)}+cos{(x)} lt -2$$$$sqrt{2} sin{(x)}lt-2$$$$sin{(x)} lt -sqrt{2}$$
For which there are no solutions where $x in mathbb{R}$.

share|cite|improve this answer

answered Jan 20 at 11:25

Peter ForemanPeter Foreman

2,7271214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  begin{align*} sin x + cos x & = sqrt{2}left(frac{1}{sqrt{2}}sin x + frac{1}{sqrt{2}}cos xright)\ & = sqrt{2}left[sin xcosleft(frac{pi}{4}right) + cos xsinleft(frac{pi}{4}right)right]\ & = sqrt{2}sinleft(x + frac{pi}{4}right) end{align*}
  $endgroup$
  – N. F. Taussig
  Jan 20 at 12:49
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

We can make a substitution to solve this problem. By letting $t=sin{(x)}+cos{(x)}$ the problem simplifies to: $$t+t^2-1gt1$$$$t^2+t-2 gt 0$$$$(t+2)(t-1)gt0$$
For which the solutions are $t gt 1$ and $t lt -2$. So we then have for $t gt 1$:$$sin{(x)}+cos{(x)} gt 1$$$$sqrt{2} sin{(x)}gt1$$$$sin{(x)} gt frac{1}{sqrt{2}}$$$$2πnlt x lt 2πn+frac{π}{2}, n in mathbb{Z}$$
Or for $t lt -2$:$$sin{(x)}+cos{(x)} lt -2$$$$sqrt{2} sin{(x)}lt-2$$$$sin{(x)} lt -sqrt{2}$$
For which there are no solutions where $x in mathbb{R}$.

share|cite|improve this answer

answered Jan 20 at 11:25

Peter ForemanPeter Foreman

2,7271214

$endgroup$

We can make a substitution to solve this problem. By letting $t=sin{(x)}+cos{(x)}$ the problem simplifies to: $$t+t^2-1gt1$$$$t^2+t-2 gt 0$$$$(t+2)(t-1)gt0$$
For which the solutions are $t gt 1$ and $t lt -2$. So we then have for $t gt 1$:$$sin{(x)}+cos{(x)} gt 1$$$$sqrt{2} sin{(x)}gt1$$$$sin{(x)} gt frac{1}{sqrt{2}}$$$$2πnlt x lt 2πn+frac{π}{2}, n in mathbb{Z}$$
Or for $t lt -2$:$$sin{(x)}+cos{(x)} lt -2$$$$sqrt{2} sin{(x)}lt-2$$$$sin{(x)} lt -sqrt{2}$$
For which there are no solutions where $x in mathbb{R}$.

share|cite|improve this answer

answered Jan 20 at 11:25

Peter ForemanPeter Foreman

2,7271214

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 11:25

Peter ForemanPeter Foreman

2,7271214

answered Jan 20 at 11:25

Peter ForemanPeter Foreman

2,7271214

answered Jan 20 at 11:25

Peter ForemanPeter Foreman

2,7271214

2,7271214

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  begin{align*} sin x + cos x & = sqrt{2}left(frac{1}{sqrt{2}}sin x + frac{1}{sqrt{2}}cos xright)\ & = sqrt{2}left[sin xcosleft(frac{pi}{4}right) + cos xsinleft(frac{pi}{4}right)right]\ & = sqrt{2}sinleft(x + frac{pi}{4}right) end{align*}
  $endgroup$
  – N. F. Taussig
  Jan 20 at 12:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  begin{align*} sin x + cos x & = sqrt{2}left(frac{1}{sqrt{2}}sin x + frac{1}{sqrt{2}}cos xright)\ & = sqrt{2}left[sin xcosleft(frac{pi}{4}right) + cos xsinleft(frac{pi}{4}right)right]\ & = sqrt{2}sinleft(x + frac{pi}{4}right) end{align*}
  $endgroup$
  – N. F. Taussig
  Jan 20 at 12:49
  

  
  
  
  

$begingroup$
begin{align*} sin x + cos x & = sqrt{2}left(frac{1}{sqrt{2}}sin x + frac{1}{sqrt{2}}cos xright)\ & = sqrt{2}left[sin xcosleft(frac{pi}{4}right) + cos xsinleft(frac{pi}{4}right)right]\ & = sqrt{2}sinleft(x + frac{pi}{4}right) end{align*}
$endgroup$
– N. F. Taussig
Jan 20 at 12:49

$begingroup$
begin{align*} sin x + cos x & = sqrt{2}left(frac{1}{sqrt{2}}sin x + frac{1}{sqrt{2}}cos xright)\ & = sqrt{2}left[sin xcosleft(frac{pi}{4}right) + cos xsinleft(frac{pi}{4}right)right]\ & = sqrt{2}sinleft(x + frac{pi}{4}right) end{align*}
$endgroup$
– N. F. Taussig
Jan 20 at 12:49

add a comment | 

0

$begingroup$

Hint: Substitute $$t=sin(x)+cos(x)$$ and square it and note that $$sin(2x)=2sin(x)cos(x)$$

share|cite|improve this answer

answered Jan 20 at 11:21

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

$endgroup$

add a comment | 

0

$begingroup$

Hint: Substitute $$t=sin(x)+cos(x)$$ and square it and note that $$sin(2x)=2sin(x)cos(x)$$

share|cite|improve this answer

answered Jan 20 at 11:21

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint: Substitute $$t=sin(x)+cos(x)$$ and square it and note that $$sin(2x)=2sin(x)cos(x)$$

share|cite|improve this answer

answered Jan 20 at 11:21

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

$endgroup$

Hint: Substitute $$t=sin(x)+cos(x)$$ and square it and note that $$sin(2x)=2sin(x)cos(x)$$

share|cite|improve this answer

answered Jan 20 at 11:21

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 11:21

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

answered Jan 20 at 11:21

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

answered Jan 20 at 11:21

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

76.8k42866

add a comment | 

add a comment | 

This page is only for reference, If you need detailed information, please check here