Solution for $(acdot x+b)cdot sin(x)+ccdot cos(x)=0$?












0












$begingroup$


As part of an engineering problem, I've been trying for a generic solution for $(acdot x+b)cdot sin(x)+ccdot cos(x)=0$



Here's something I tried:



$(a⋅x+b)⋅tan(x)+c=0$, when $cos(x)neq 0$
$$tan(x)=frac{-c}{a⋅x+b}$$
It sort of looking like this, when, say, $a=2, b=3, c=5$:a=2, b=3, c=5



Any chance that can be solved without using numerical methods?



EDIT: I was looking for solutions in the form of $x approx f(a,b,c) $, preferably finding solution in any range, because $a$, $b$, and $c$ can take wierd values. It's used for a program where it will not be trivial to solve problem procedurally.






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think not possible!
    $endgroup$
    – gimusi
    Dec 13 '18 at 15:06










  • $begingroup$
    No. See the similar type of question Numerical solution to $x = tan (x)$ for some info on how to get numerical solutions
    $endgroup$
    – Winther
    Dec 13 '18 at 15:10








  • 2




    $begingroup$
    No chance. Rewrite as $cot x=px+q$ and find the intersections between the line and the vertical asymptotes to get starting approximations.
    $endgroup$
    – Yves Daoust
    Dec 13 '18 at 15:11
















0












$begingroup$


As part of an engineering problem, I've been trying for a generic solution for $(acdot x+b)cdot sin(x)+ccdot cos(x)=0$



Here's something I tried:



$(a⋅x+b)⋅tan(x)+c=0$, when $cos(x)neq 0$
$$tan(x)=frac{-c}{a⋅x+b}$$
It sort of looking like this, when, say, $a=2, b=3, c=5$:a=2, b=3, c=5



Any chance that can be solved without using numerical methods?



EDIT: I was looking for solutions in the form of $x approx f(a,b,c) $, preferably finding solution in any range, because $a$, $b$, and $c$ can take wierd values. It's used for a program where it will not be trivial to solve problem procedurally.






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think not possible!
    $endgroup$
    – gimusi
    Dec 13 '18 at 15:06










  • $begingroup$
    No. See the similar type of question Numerical solution to $x = tan (x)$ for some info on how to get numerical solutions
    $endgroup$
    – Winther
    Dec 13 '18 at 15:10








  • 2




    $begingroup$
    No chance. Rewrite as $cot x=px+q$ and find the intersections between the line and the vertical asymptotes to get starting approximations.
    $endgroup$
    – Yves Daoust
    Dec 13 '18 at 15:11














0












0








0





$begingroup$


As part of an engineering problem, I've been trying for a generic solution for $(acdot x+b)cdot sin(x)+ccdot cos(x)=0$



Here's something I tried:



$(a⋅x+b)⋅tan(x)+c=0$, when $cos(x)neq 0$
$$tan(x)=frac{-c}{a⋅x+b}$$
It sort of looking like this, when, say, $a=2, b=3, c=5$:a=2, b=3, c=5



Any chance that can be solved without using numerical methods?



EDIT: I was looking for solutions in the form of $x approx f(a,b,c) $, preferably finding solution in any range, because $a$, $b$, and $c$ can take wierd values. It's used for a program where it will not be trivial to solve problem procedurally.






trigonometry







share|cite|improve this question











$endgroup$




As part of an engineering problem, I've been trying for a generic solution for $(acdot x+b)cdot sin(x)+ccdot cos(x)=0$



Here's something I tried:



$(a⋅x+b)⋅tan(x)+c=0$, when $cos(x)neq 0$
$$tan(x)=frac{-c}{a⋅x+b}$$
It sort of looking like this, when, say, $a=2, b=3, c=5$:a=2, b=3, c=5



Any chance that can be solved without using numerical methods?



EDIT: I was looking for solutions in the form of $x approx f(a,b,c) $, preferably finding solution in any range, because $a$, $b$, and $c$ can take wierd values. It's used for a program where it will not be trivial to solve problem procedurally.






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 18:45







zyc

















asked Dec 13 '18 at 14:58









zyczyc

1114




1114












  • $begingroup$
    I think not possible!
    $endgroup$
    – gimusi
    Dec 13 '18 at 15:06










  • $begingroup$
    No. See the similar type of question Numerical solution to $x = tan (x)$ for some info on how to get numerical solutions
    $endgroup$
    – Winther
    Dec 13 '18 at 15:10








  • 2




    $begingroup$
    No chance. Rewrite as $cot x=px+q$ and find the intersections between the line and the vertical asymptotes to get starting approximations.
    $endgroup$
    – Yves Daoust
    Dec 13 '18 at 15:11


















  • $begingroup$
    I think not possible!
    $endgroup$
    – gimusi
    Dec 13 '18 at 15:06










  • $begingroup$
    No. See the similar type of question Numerical solution to $x = tan (x)$ for some info on how to get numerical solutions
    $endgroup$
    – Winther
    Dec 13 '18 at 15:10








  • 2




    $begingroup$
    No chance. Rewrite as $cot x=px+q$ and find the intersections between the line and the vertical asymptotes to get starting approximations.
    $endgroup$
    – Yves Daoust
    Dec 13 '18 at 15:11
















$begingroup$
I think not possible!
$endgroup$
– gimusi
Dec 13 '18 at 15:06




$begingroup$
I think not possible!
$endgroup$
– gimusi
Dec 13 '18 at 15:06












$begingroup$
No. See the similar type of question Numerical solution to $x = tan (x)$ for some info on how to get numerical solutions
$endgroup$
– Winther
Dec 13 '18 at 15:10






$begingroup$
No. See the similar type of question Numerical solution to $x = tan (x)$ for some info on how to get numerical solutions
$endgroup$
– Winther
Dec 13 '18 at 15:10






2




2




$begingroup$
No chance. Rewrite as $cot x=px+q$ and find the intersections between the line and the vertical asymptotes to get starting approximations.
$endgroup$
– Yves Daoust
Dec 13 '18 at 15:11




$begingroup$
No chance. Rewrite as $cot x=px+q$ and find the intersections between the line and the vertical asymptotes to get starting approximations.
$endgroup$
– Yves Daoust
Dec 13 '18 at 15:11










2 Answers
2






active

oldest

votes


















1












$begingroup$

There's no closed form solution for that equation, so numerical methods are required.



That said, you can come up with estimates for the solutions towards $pm infty$.
For instance, as $nrightarrow +infty$, if you write the solutions as $x_n=npi + u_n$, with $u_nin (-frac pi 2, frac pi 2)$, then
$$tan x_n = tan u_n = frac {-c}{ax_n+b}tag{1}=-frac c {anpi + a u_n + b}$$
Because the right-hand side of that equation is equivalent to $-frac c {api n}$, we must have that $tan u_nrightarrow 0$, which means that $u_nrightarrow 0$.
Plugging that information back into (1) yields
$$u_n=-frac c {api n} + v_n$$ where $v_n = o(frac 1 n)$.
And you can now plug this back, again, into (1), and find an estimate for $v_n$. And you keep doing that ad lib to find higher order estimates in the solution
$$x_n = npi -frac c {api n} + ....$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'll try that to see if I can rewrite it as $x_n = f(a,b,c)$. I'm not a mathematician, so I'm not really sure.
    $endgroup$
    – zyc
    Dec 13 '18 at 16:22





















1












$begingroup$

Besides numerical solutions for particular parameter values, you could try series solutions. Thus when $c$ is small compared to $a$ and $b$, the solution near $x=0$ is



$$ -frac{b}{a} + cotleft(frac{b}{a}right) frac{c}{a} + cotleft(frac{b}{a}right) csc^2 left(frac{b}{a}right) frac{c^2}{a^2}
+ left(cot^3left(frac{b}{a}right) + cotleft(frac{b}{a}right) sec^4 left(frac{b}{a}right)right) frac{c^3}{a^3} + ldots $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $$left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot sec^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=1right )+left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot csc^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=0right )$$
    $endgroup$
    – zyc
    Dec 13 '18 at 16:03










  • $begingroup$
    Exactly what I need! Is there a tutorial on how to derive that?
    $endgroup$
    – zyc
    Dec 13 '18 at 16:04











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

There's no closed form solution for that equation, so numerical methods are required.



That said, you can come up with estimates for the solutions towards $pm infty$.
For instance, as $nrightarrow +infty$, if you write the solutions as $x_n=npi + u_n$, with $u_nin (-frac pi 2, frac pi 2)$, then
$$tan x_n = tan u_n = frac {-c}{ax_n+b}tag{1}=-frac c {anpi + a u_n + b}$$
Because the right-hand side of that equation is equivalent to $-frac c {api n}$, we must have that $tan u_nrightarrow 0$, which means that $u_nrightarrow 0$.
Plugging that information back into (1) yields
$$u_n=-frac c {api n} + v_n$$ where $v_n = o(frac 1 n)$.
And you can now plug this back, again, into (1), and find an estimate for $v_n$. And you keep doing that ad lib to find higher order estimates in the solution
$$x_n = npi -frac c {api n} + ....$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'll try that to see if I can rewrite it as $x_n = f(a,b,c)$. I'm not a mathematician, so I'm not really sure.
    $endgroup$
    – zyc
    Dec 13 '18 at 16:22


















1












$begingroup$

There's no closed form solution for that equation, so numerical methods are required.



That said, you can come up with estimates for the solutions towards $pm infty$.
For instance, as $nrightarrow +infty$, if you write the solutions as $x_n=npi + u_n$, with $u_nin (-frac pi 2, frac pi 2)$, then
$$tan x_n = tan u_n = frac {-c}{ax_n+b}tag{1}=-frac c {anpi + a u_n + b}$$
Because the right-hand side of that equation is equivalent to $-frac c {api n}$, we must have that $tan u_nrightarrow 0$, which means that $u_nrightarrow 0$.
Plugging that information back into (1) yields
$$u_n=-frac c {api n} + v_n$$ where $v_n = o(frac 1 n)$.
And you can now plug this back, again, into (1), and find an estimate for $v_n$. And you keep doing that ad lib to find higher order estimates in the solution
$$x_n = npi -frac c {api n} + ....$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'll try that to see if I can rewrite it as $x_n = f(a,b,c)$. I'm not a mathematician, so I'm not really sure.
    $endgroup$
    – zyc
    Dec 13 '18 at 16:22
















1












1








1





$begingroup$

There's no closed form solution for that equation, so numerical methods are required.



That said, you can come up with estimates for the solutions towards $pm infty$.
For instance, as $nrightarrow +infty$, if you write the solutions as $x_n=npi + u_n$, with $u_nin (-frac pi 2, frac pi 2)$, then
$$tan x_n = tan u_n = frac {-c}{ax_n+b}tag{1}=-frac c {anpi + a u_n + b}$$
Because the right-hand side of that equation is equivalent to $-frac c {api n}$, we must have that $tan u_nrightarrow 0$, which means that $u_nrightarrow 0$.
Plugging that information back into (1) yields
$$u_n=-frac c {api n} + v_n$$ where $v_n = o(frac 1 n)$.
And you can now plug this back, again, into (1), and find an estimate for $v_n$. And you keep doing that ad lib to find higher order estimates in the solution
$$x_n = npi -frac c {api n} + ....$$






share|cite|improve this answer









$endgroup$



There's no closed form solution for that equation, so numerical methods are required.



That said, you can come up with estimates for the solutions towards $pm infty$.
For instance, as $nrightarrow +infty$, if you write the solutions as $x_n=npi + u_n$, with $u_nin (-frac pi 2, frac pi 2)$, then
$$tan x_n = tan u_n = frac {-c}{ax_n+b}tag{1}=-frac c {anpi + a u_n + b}$$
Because the right-hand side of that equation is equivalent to $-frac c {api n}$, we must have that $tan u_nrightarrow 0$, which means that $u_nrightarrow 0$.
Plugging that information back into (1) yields
$$u_n=-frac c {api n} + v_n$$ where $v_n = o(frac 1 n)$.
And you can now plug this back, again, into (1), and find an estimate for $v_n$. And you keep doing that ad lib to find higher order estimates in the solution
$$x_n = npi -frac c {api n} + ....$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 15:30









Stefan LafonStefan Lafon

2,24019




2,24019












  • $begingroup$
    I'll try that to see if I can rewrite it as $x_n = f(a,b,c)$. I'm not a mathematician, so I'm not really sure.
    $endgroup$
    – zyc
    Dec 13 '18 at 16:22




















  • $begingroup$
    I'll try that to see if I can rewrite it as $x_n = f(a,b,c)$. I'm not a mathematician, so I'm not really sure.
    $endgroup$
    – zyc
    Dec 13 '18 at 16:22


















$begingroup$
I'll try that to see if I can rewrite it as $x_n = f(a,b,c)$. I'm not a mathematician, so I'm not really sure.
$endgroup$
– zyc
Dec 13 '18 at 16:22






$begingroup$
I'll try that to see if I can rewrite it as $x_n = f(a,b,c)$. I'm not a mathematician, so I'm not really sure.
$endgroup$
– zyc
Dec 13 '18 at 16:22













1












$begingroup$

Besides numerical solutions for particular parameter values, you could try series solutions. Thus when $c$ is small compared to $a$ and $b$, the solution near $x=0$ is



$$ -frac{b}{a} + cotleft(frac{b}{a}right) frac{c}{a} + cotleft(frac{b}{a}right) csc^2 left(frac{b}{a}right) frac{c^2}{a^2}
+ left(cot^3left(frac{b}{a}right) + cotleft(frac{b}{a}right) sec^4 left(frac{b}{a}right)right) frac{c^3}{a^3} + ldots $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $$left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot sec^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=1right )+left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot csc^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=0right )$$
    $endgroup$
    – zyc
    Dec 13 '18 at 16:03










  • $begingroup$
    Exactly what I need! Is there a tutorial on how to derive that?
    $endgroup$
    – zyc
    Dec 13 '18 at 16:04
















1












$begingroup$

Besides numerical solutions for particular parameter values, you could try series solutions. Thus when $c$ is small compared to $a$ and $b$, the solution near $x=0$ is



$$ -frac{b}{a} + cotleft(frac{b}{a}right) frac{c}{a} + cotleft(frac{b}{a}right) csc^2 left(frac{b}{a}right) frac{c^2}{a^2}
+ left(cot^3left(frac{b}{a}right) + cotleft(frac{b}{a}right) sec^4 left(frac{b}{a}right)right) frac{c^3}{a^3} + ldots $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $$left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot sec^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=1right )+left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot csc^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=0right )$$
    $endgroup$
    – zyc
    Dec 13 '18 at 16:03










  • $begingroup$
    Exactly what I need! Is there a tutorial on how to derive that?
    $endgroup$
    – zyc
    Dec 13 '18 at 16:04














1












1








1





$begingroup$

Besides numerical solutions for particular parameter values, you could try series solutions. Thus when $c$ is small compared to $a$ and $b$, the solution near $x=0$ is



$$ -frac{b}{a} + cotleft(frac{b}{a}right) frac{c}{a} + cotleft(frac{b}{a}right) csc^2 left(frac{b}{a}right) frac{c^2}{a^2}
+ left(cot^3left(frac{b}{a}right) + cotleft(frac{b}{a}right) sec^4 left(frac{b}{a}right)right) frac{c^3}{a^3} + ldots $$






share|cite|improve this answer









$endgroup$



Besides numerical solutions for particular parameter values, you could try series solutions. Thus when $c$ is small compared to $a$ and $b$, the solution near $x=0$ is



$$ -frac{b}{a} + cotleft(frac{b}{a}right) frac{c}{a} + cotleft(frac{b}{a}right) csc^2 left(frac{b}{a}right) frac{c^2}{a^2}
+ left(cot^3left(frac{b}{a}right) + cotleft(frac{b}{a}right) sec^4 left(frac{b}{a}right)right) frac{c^3}{a^3} + ldots $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 15:31









Robert IsraelRobert Israel

325k23214468




325k23214468












  • $begingroup$
    $$left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot sec^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=1right )+left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot csc^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=0right )$$
    $endgroup$
    – zyc
    Dec 13 '18 at 16:03










  • $begingroup$
    Exactly what I need! Is there a tutorial on how to derive that?
    $endgroup$
    – zyc
    Dec 13 '18 at 16:04


















  • $begingroup$
    $$left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot sec^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=1right )+left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot csc^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=0right )$$
    $endgroup$
    – zyc
    Dec 13 '18 at 16:03










  • $begingroup$
    Exactly what I need! Is there a tutorial on how to derive that?
    $endgroup$
    – zyc
    Dec 13 '18 at 16:04
















$begingroup$
$$left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot sec^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=1right )+left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot csc^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=0right )$$
$endgroup$
– zyc
Dec 13 '18 at 16:03




$begingroup$
$$left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot sec^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=1right )+left ( sum_{i=1}^{infty }left ( cot^{i}frac{b}{a}+cotfrac{b}{a}cdot csc^{2left ( i-1 right )}frac{b}{a} right )cdot frac{c^{i-1}}{a^{i-1}}mid imod 2=0right )$$
$endgroup$
– zyc
Dec 13 '18 at 16:03












$begingroup$
Exactly what I need! Is there a tutorial on how to derive that?
$endgroup$
– zyc
Dec 13 '18 at 16:04




$begingroup$
Exactly what I need! Is there a tutorial on how to derive that?
$endgroup$
– zyc
Dec 13 '18 at 16:04


















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Bundesstraße 106

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Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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