Fdtxjr

This word with a lot of past tenses

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If the equation $sin^2x-asin x+b=0$ has only one solution in $(0,pi)$, then what is the range of $b$?

Show that $sin^2 x- 6sin x-5=0$ has more than one real solutionsin (x) + cos (x) = 0. Why this equation has only one solution set?If $ -3left(x-lfloor x rfloor right)^2+2(x-lfloor x rfloor )+a^2=0$ has no integral solution, then $a$ isFind the smallest positive number $p$ for which the equation $cos(psin x)=sin(p cos x)$ has a solution $xin[0,2pi].$General Solution of the equation $sin^2015(phi)+cos^2015(phi) = 1$Find the general solution to this a equation.Find range of 'a' for which the eqn. has at least one real solution.Prove there is only one solution and find only solution of $1 + sqrt 2 x +sin x - cos x = 0 $Real solution of $(cos x -sin x)cdot bigg(2tan x+frac1cos xbigg)+2=0.$value of $a,b$ in trigonometric expression

1

$begingroup$

If the equation $sin^2(x)-asin(x)+b=0$ has only one solution in $(0,pi)$, then what is the range of $b$?

What I try:
$$displaystyle sin x=fracapm sqrta^2-4b2inbigg(0,1bigg]$$

for one real solution $a^2=4b$.

How do I solve it? Help me, please.

trigonometry quadratics

share|cite|improve this question

edited Mar 12 at 11:58

Blue

49.1k870156

asked Mar 12 at 11:34

jackyjacky

1,186715

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: When does the expression (with the $pm$) have exactly one outcome?
  $endgroup$
  – Matti P.
  Mar 12 at 11:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: There is only one value $y$ such that $x in (0,pi): sin(x)=y$ has cardinality $1$.
  $endgroup$
  – Robert Israel
  Mar 12 at 12:19
  

  
  
  
  

add a comment | 

1

$begingroup$

If the equation $sin^2(x)-asin(x)+b=0$ has only one solution in $(0,pi)$, then what is the range of $b$?

What I try:
$$displaystyle sin x=fracapm sqrta^2-4b2inbigg(0,1bigg]$$

for one real solution $a^2=4b$.

How do I solve it? Help me, please.

trigonometry quadratics

share|cite|improve this question

edited Mar 12 at 11:58

Blue

49.1k870156

asked Mar 12 at 11:34

jackyjacky

1,186715

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: When does the expression (with the $pm$) have exactly one outcome?
  $endgroup$
  – Matti P.
  Mar 12 at 11:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: There is only one value $y$ such that $x in (0,pi): sin(x)=y$ has cardinality $1$.
  $endgroup$
  – Robert Israel
  Mar 12 at 12:19
  

  
  
  
  

add a comment | 

$begingroup$

If the equation $sin^2(x)-asin(x)+b=0$ has only one solution in $(0,pi)$, then what is the range of $b$?

What I try:
$$displaystyle sin x=fracapm sqrta^2-4b2inbigg(0,1bigg]$$

for one real solution $a^2=4b$.

How do I solve it? Help me, please.

trigonometry quadratics

share|cite|improve this question

edited Mar 12 at 11:58

Blue

49.1k870156

asked Mar 12 at 11:34

jackyjacky

1,186715

$endgroup$

share|cite|improve this question

edited Mar 12 at 11:58

Blue

49.1k870156

asked Mar 12 at 11:34

jackyjacky

1,186715

share|cite|improve this question

edited Mar 12 at 11:58

Blue

49.1k870156

asked Mar 12 at 11:34

jackyjacky

1,186715

edited Mar 12 at 11:58

Blue

49.1k870156

edited Mar 12 at 11:58

Blue

49.1k870156

asked Mar 12 at 11:34

jackyjacky

1,186715

asked Mar 12 at 11:34

jackyjacky

1,186715

3 Answers
3

active

oldest

votes

2

$begingroup$

If $xin (0,pi)$ solves the equation then $pi-xin (0,pi)$ also solves the equation since it holds $sin x = sin(pi-x)$. By the uniqueness of solution on $(0,pi)$, we have $x=pi-x$, i.e. $x=fracpi2$. Hence $sin(frac pi2)=1$ solves the quadratic equation, which implies that $$
sin^2(x)-asin x+b=(sin x-1)(sin x-b).
$$ In order that $sin x = b$ has no solution on $(0,pi)$ other than $x=fracpi2$, it must be that $$
bge 1 text or ble 0.
$$

share|cite|improve this answer

answered Mar 12 at 11:49

SongSong

18k21449

$endgroup$

add a comment | 

1

$begingroup$

Hint; Since we have $$|sin(x)|le 1$$ you have to solve
$$left|frac12left(apmsqrta^2-4bright)right|le 1$$

share|cite|improve this answer

answered Mar 12 at 11:56

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.9k42866

$endgroup$

add a comment | 

1

$begingroup$

To $x$ to have only one solution in the range $(0,pi)$, then $x$ must be $piover 2$, according to the unit circle for trigonometry.

Since, $sinfracpi2 = 1$,
$$sin^2 x - asin x +b = 1-a+b =0$$
$$implies a-b=1 totexteq.1 $$
From what you have tried we get,
$$a^2 =4b to texteq.2 $$
Thus, to satisfy eqs. 1&2 is,
From eq.1 , we get,
$$a=1+b$$
Subtituting into eq.2,
$$(1+b)^2=4b$$
$$1+2b+b^2=4b$$
$$(b-1)^2=0$$
$$therefore b=1$$
and $a=2$

share|cite|improve this answer

edited Mar 12 at 13:21

answered Mar 12 at 13:08

rashrash

50914

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  first equation is invalid. It should be $1-a+b=0$
  $endgroup$
  – Takahiro Waki
  Mar 12 at 13:18
  

  
  
  
  

add a comment | 

Your Answer

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2

$begingroup$

If $xin (0,pi)$ solves the equation then $pi-xin (0,pi)$ also solves the equation since it holds $sin x = sin(pi-x)$. By the uniqueness of solution on $(0,pi)$, we have $x=pi-x$, i.e. $x=fracpi2$. Hence $sin(frac pi2)=1$ solves the quadratic equation, which implies that $$
sin^2(x)-asin x+b=(sin x-1)(sin x-b).
$$ In order that $sin x = b$ has no solution on $(0,pi)$ other than $x=fracpi2$, it must be that $$
bge 1 text or ble 0.
$$

share|cite|improve this answer

answered Mar 12 at 11:49

SongSong

18k21449

$endgroup$

add a comment | 

2

$begingroup$

If $xin (0,pi)$ solves the equation then $pi-xin (0,pi)$ also solves the equation since it holds $sin x = sin(pi-x)$. By the uniqueness of solution on $(0,pi)$, we have $x=pi-x$, i.e. $x=fracpi2$. Hence $sin(frac pi2)=1$ solves the quadratic equation, which implies that $$
sin^2(x)-asin x+b=(sin x-1)(sin x-b).
$$ In order that $sin x = b$ has no solution on $(0,pi)$ other than $x=fracpi2$, it must be that $$
bge 1 text or ble 0.
$$

share|cite|improve this answer

answered Mar 12 at 11:49

SongSong

18k21449

$endgroup$

add a comment | 

$begingroup$

If $xin (0,pi)$ solves the equation then $pi-xin (0,pi)$ also solves the equation since it holds $sin x = sin(pi-x)$. By the uniqueness of solution on $(0,pi)$, we have $x=pi-x$, i.e. $x=fracpi2$. Hence $sin(frac pi2)=1$ solves the quadratic equation, which implies that $$
sin^2(x)-asin x+b=(sin x-1)(sin x-b).
$$ In order that $sin x = b$ has no solution on $(0,pi)$ other than $x=fracpi2$, it must be that $$
bge 1 text or ble 0.
$$

share|cite|improve this answer

answered Mar 12 at 11:49

SongSong

18k21449

$endgroup$

share|cite|improve this answer

answered Mar 12 at 11:49

SongSong

18k21449

answered Mar 12 at 11:49

SongSong

18k21449

answered Mar 12 at 11:49

SongSong

18k21449

1

$begingroup$

Hint; Since we have $$|sin(x)|le 1$$ you have to solve
$$left|frac12left(apmsqrta^2-4bright)right|le 1$$

share|cite|improve this answer

answered Mar 12 at 11:56

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.9k42866

$endgroup$

add a comment | 

1

$begingroup$

Hint; Since we have $$|sin(x)|le 1$$ you have to solve
$$left|frac12left(apmsqrta^2-4bright)right|le 1$$

share|cite|improve this answer

answered Mar 12 at 11:56

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.9k42866

$endgroup$

add a comment | 

$begingroup$

Hint; Since we have $$|sin(x)|le 1$$ you have to solve
$$left|frac12left(apmsqrta^2-4bright)right|le 1$$

share|cite|improve this answer

answered Mar 12 at 11:56

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.9k42866

$endgroup$

share|cite|improve this answer

answered Mar 12 at 11:56

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.9k42866

answered Mar 12 at 11:56

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.9k42866

answered Mar 12 at 11:56

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.9k42866

1

$begingroup$

To $x$ to have only one solution in the range $(0,pi)$, then $x$ must be $piover 2$, according to the unit circle for trigonometry.

Since, $sinfracpi2 = 1$,
$$sin^2 x - asin x +b = 1-a+b =0$$
$$implies a-b=1 totexteq.1 $$
From what you have tried we get,
$$a^2 =4b to texteq.2 $$
Thus, to satisfy eqs. 1&2 is,
From eq.1 , we get,
$$a=1+b$$
Subtituting into eq.2,
$$(1+b)^2=4b$$
$$1+2b+b^2=4b$$
$$(b-1)^2=0$$
$$therefore b=1$$
and $a=2$

share|cite|improve this answer

edited Mar 12 at 13:21

answered Mar 12 at 13:08

rashrash

50914

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  first equation is invalid. It should be $1-a+b=0$
  $endgroup$
  – Takahiro Waki
  Mar 12 at 13:18
  

  
  
  
  

add a comment | 

1

$begingroup$

To $x$ to have only one solution in the range $(0,pi)$, then $x$ must be $piover 2$, according to the unit circle for trigonometry.

Since, $sinfracpi2 = 1$,
$$sin^2 x - asin x +b = 1-a+b =0$$
$$implies a-b=1 totexteq.1 $$
From what you have tried we get,
$$a^2 =4b to texteq.2 $$
Thus, to satisfy eqs. 1&2 is,
From eq.1 , we get,
$$a=1+b$$
Subtituting into eq.2,
$$(1+b)^2=4b$$
$$1+2b+b^2=4b$$
$$(b-1)^2=0$$
$$therefore b=1$$
and $a=2$

share|cite|improve this answer

edited Mar 12 at 13:21

answered Mar 12 at 13:08

rashrash

50914

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  first equation is invalid. It should be $1-a+b=0$
  $endgroup$
  – Takahiro Waki
  Mar 12 at 13:18
  

  
  
  
  

add a comment | 

$begingroup$

To $x$ to have only one solution in the range $(0,pi)$, then $x$ must be $piover 2$, according to the unit circle for trigonometry.

Since, $sinfracpi2 = 1$,
$$sin^2 x - asin x +b = 1-a+b =0$$
$$implies a-b=1 totexteq.1 $$
From what you have tried we get,
$$a^2 =4b to texteq.2 $$
Thus, to satisfy eqs. 1&2 is,
From eq.1 , we get,
$$a=1+b$$
Subtituting into eq.2,
$$(1+b)^2=4b$$
$$1+2b+b^2=4b$$
$$(b-1)^2=0$$
$$therefore b=1$$
and $a=2$

share|cite|improve this answer

edited Mar 12 at 13:21

answered Mar 12 at 13:08

rashrash

50914

$endgroup$

share|cite|improve this answer

edited Mar 12 at 13:21

answered Mar 12 at 13:08

rashrash

50914

answered Mar 12 at 13:08

rashrash

50914

answered Mar 12 at 13:08

rashrash

50914