Calculation of $x$ in $tan^{-1}left(frac{x}{1-x^2}right)+tan^{-1}left(frac{1}{x^3}right) = frac{3pi}{4}$












1














The no. of real values of $x$ satisfying $displaystyle tan^{-1}left(frac{x}{1-x^2}right)+tan^{-1}left(frac{1}{x^3}right) = frac{3pi}{4}$



options ::



(a) $0$



(b) $1$



(c) $2$



(d) Infinitely many



My Try:: Using The formula $tan^{-1}A+tan^{-1}B = tan^{-1}left(frac{A+B}{1-AB}right)$



So $displaystyle tan^{-1}left(frac{frac{x}{1-x^2}+frac{1}{x^3}}{1-frac{x}{x^3.(1-x^2)}}right) = frac{3pi}{4}$



So $displaystyle frac{x^4+1-x^2}{x^3-x^5-x}= -1$



So $(x-1)(x^4-x+1) = 0$



$x = 1$ and $x^4 -x+1 = 0$



So Iam getting only one value of $x$ which is $x=1$ and How can i calculate any real value of $x$ exists from $x^4-x+1 = 0$



OR any other method by using we can solve this Question



Thanks






trigonometry







share|cite|improve this question
























  • The equation $x^4-x+1=0$ has no real solution.
    – Hanul Jeon
    Jan 26 '13 at 10:28










  • @juantheron I noticed you've asked several question on MSE, but you've never accepted an answer. You should accept your favorite answer if it is satisfactory.
    – Git Gud
    Jan 26 '13 at 10:46










  • O sorry Git Gud ....
    – juantheron
    Jan 26 '13 at 11:15
















1














The no. of real values of $x$ satisfying $displaystyle tan^{-1}left(frac{x}{1-x^2}right)+tan^{-1}left(frac{1}{x^3}right) = frac{3pi}{4}$



options ::



(a) $0$



(b) $1$



(c) $2$



(d) Infinitely many



My Try:: Using The formula $tan^{-1}A+tan^{-1}B = tan^{-1}left(frac{A+B}{1-AB}right)$



So $displaystyle tan^{-1}left(frac{frac{x}{1-x^2}+frac{1}{x^3}}{1-frac{x}{x^3.(1-x^2)}}right) = frac{3pi}{4}$



So $displaystyle frac{x^4+1-x^2}{x^3-x^5-x}= -1$



So $(x-1)(x^4-x+1) = 0$



$x = 1$ and $x^4 -x+1 = 0$



So Iam getting only one value of $x$ which is $x=1$ and How can i calculate any real value of $x$ exists from $x^4-x+1 = 0$



OR any other method by using we can solve this Question



Thanks






trigonometry







share|cite|improve this question
























  • The equation $x^4-x+1=0$ has no real solution.
    – Hanul Jeon
    Jan 26 '13 at 10:28










  • @juantheron I noticed you've asked several question on MSE, but you've never accepted an answer. You should accept your favorite answer if it is satisfactory.
    – Git Gud
    Jan 26 '13 at 10:46










  • O sorry Git Gud ....
    – juantheron
    Jan 26 '13 at 11:15














1












1








1







The no. of real values of $x$ satisfying $displaystyle tan^{-1}left(frac{x}{1-x^2}right)+tan^{-1}left(frac{1}{x^3}right) = frac{3pi}{4}$



options ::



(a) $0$



(b) $1$



(c) $2$



(d) Infinitely many



My Try:: Using The formula $tan^{-1}A+tan^{-1}B = tan^{-1}left(frac{A+B}{1-AB}right)$



So $displaystyle tan^{-1}left(frac{frac{x}{1-x^2}+frac{1}{x^3}}{1-frac{x}{x^3.(1-x^2)}}right) = frac{3pi}{4}$



So $displaystyle frac{x^4+1-x^2}{x^3-x^5-x}= -1$



So $(x-1)(x^4-x+1) = 0$



$x = 1$ and $x^4 -x+1 = 0$



So Iam getting only one value of $x$ which is $x=1$ and How can i calculate any real value of $x$ exists from $x^4-x+1 = 0$



OR any other method by using we can solve this Question



Thanks






trigonometry







share|cite|improve this question















The no. of real values of $x$ satisfying $displaystyle tan^{-1}left(frac{x}{1-x^2}right)+tan^{-1}left(frac{1}{x^3}right) = frac{3pi}{4}$



options ::



(a) $0$



(b) $1$



(c) $2$



(d) Infinitely many



My Try:: Using The formula $tan^{-1}A+tan^{-1}B = tan^{-1}left(frac{A+B}{1-AB}right)$



So $displaystyle tan^{-1}left(frac{frac{x}{1-x^2}+frac{1}{x^3}}{1-frac{x}{x^3.(1-x^2)}}right) = frac{3pi}{4}$



So $displaystyle frac{x^4+1-x^2}{x^3-x^5-x}= -1$



So $(x-1)(x^4-x+1) = 0$



$x = 1$ and $x^4 -x+1 = 0$



So Iam getting only one value of $x$ which is $x=1$ and How can i calculate any real value of $x$ exists from $x^4-x+1 = 0$



OR any other method by using we can solve this Question



Thanks






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 15:04









Martin Sleziak

44.6k7115270




44.6k7115270










asked Jan 26 '13 at 10:23









juantheron

34k1145140




34k1145140












  • The equation $x^4-x+1=0$ has no real solution.
    – Hanul Jeon
    Jan 26 '13 at 10:28










  • @juantheron I noticed you've asked several question on MSE, but you've never accepted an answer. You should accept your favorite answer if it is satisfactory.
    – Git Gud
    Jan 26 '13 at 10:46










  • O sorry Git Gud ....
    – juantheron
    Jan 26 '13 at 11:15


















  • The equation $x^4-x+1=0$ has no real solution.
    – Hanul Jeon
    Jan 26 '13 at 10:28










  • @juantheron I noticed you've asked several question on MSE, but you've never accepted an answer. You should accept your favorite answer if it is satisfactory.
    – Git Gud
    Jan 26 '13 at 10:46










  • O sorry Git Gud ....
    – juantheron
    Jan 26 '13 at 11:15
















The equation $x^4-x+1=0$ has no real solution.
– Hanul Jeon
Jan 26 '13 at 10:28




The equation $x^4-x+1=0$ has no real solution.
– Hanul Jeon
Jan 26 '13 at 10:28












@juantheron I noticed you've asked several question on MSE, but you've never accepted an answer. You should accept your favorite answer if it is satisfactory.
– Git Gud
Jan 26 '13 at 10:46




@juantheron I noticed you've asked several question on MSE, but you've never accepted an answer. You should accept your favorite answer if it is satisfactory.
– Git Gud
Jan 26 '13 at 10:46












O sorry Git Gud ....
– juantheron
Jan 26 '13 at 11:15




O sorry Git Gud ....
– juantheron
Jan 26 '13 at 11:15










1 Answer
1






active

oldest

votes


















1















  1. If $|x|<1$, then $x^4+1>1>x$. If $|x|geq 1$, then $x^4+1>x^4geq x$. In both cases
    $$
    x^4-x+1>0
    $$
    so the equation $x^4-x+1=0$ have no real solutions.


  2. Note that $x=1$ is not a solution of original equation.







share|cite|improve this answer





















  • Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $displaystyle tan^{-1}(infty)+tan^{-1}(1) = frac{pi}{2}+frac{pi}{4} = frac{3pi}{4}$ which is exactly equaal to $R.H.S$
    – juantheron
    Jan 26 '13 at 10:46








  • 2




    Expression $tan^{-1}(infty)$ makes no sense. Even if you are talking about limits it is not ok because $$limlimits_{xto 1+}tan^{-1}frac{x}{1-x^2}=frac{pi}{2}$$$$limlimits_{xto 1-}tan^{-1}frac{x}{1-x^2}=-frac{pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $frac{x}{1-x^2}$
    – Norbert
    Jan 26 '13 at 10:51












  • Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a)
    – juantheron
    Jan 26 '13 at 11:16













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f287220%2fcalculation-of-x-in-tan-1-left-fracx1-x2-right-tan-1-left-fra%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1















  1. If $|x|<1$, then $x^4+1>1>x$. If $|x|geq 1$, then $x^4+1>x^4geq x$. In both cases
    $$
    x^4-x+1>0
    $$
    so the equation $x^4-x+1=0$ have no real solutions.


  2. Note that $x=1$ is not a solution of original equation.







share|cite|improve this answer





















  • Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $displaystyle tan^{-1}(infty)+tan^{-1}(1) = frac{pi}{2}+frac{pi}{4} = frac{3pi}{4}$ which is exactly equaal to $R.H.S$
    – juantheron
    Jan 26 '13 at 10:46








  • 2




    Expression $tan^{-1}(infty)$ makes no sense. Even if you are talking about limits it is not ok because $$limlimits_{xto 1+}tan^{-1}frac{x}{1-x^2}=frac{pi}{2}$$$$limlimits_{xto 1-}tan^{-1}frac{x}{1-x^2}=-frac{pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $frac{x}{1-x^2}$
    – Norbert
    Jan 26 '13 at 10:51












  • Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a)
    – juantheron
    Jan 26 '13 at 11:16


















1















  1. If $|x|<1$, then $x^4+1>1>x$. If $|x|geq 1$, then $x^4+1>x^4geq x$. In both cases
    $$
    x^4-x+1>0
    $$
    so the equation $x^4-x+1=0$ have no real solutions.


  2. Note that $x=1$ is not a solution of original equation.







share|cite|improve this answer





















  • Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $displaystyle tan^{-1}(infty)+tan^{-1}(1) = frac{pi}{2}+frac{pi}{4} = frac{3pi}{4}$ which is exactly equaal to $R.H.S$
    – juantheron
    Jan 26 '13 at 10:46








  • 2




    Expression $tan^{-1}(infty)$ makes no sense. Even if you are talking about limits it is not ok because $$limlimits_{xto 1+}tan^{-1}frac{x}{1-x^2}=frac{pi}{2}$$$$limlimits_{xto 1-}tan^{-1}frac{x}{1-x^2}=-frac{pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $frac{x}{1-x^2}$
    – Norbert
    Jan 26 '13 at 10:51












  • Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a)
    – juantheron
    Jan 26 '13 at 11:16
















1












1








1







  1. If $|x|<1$, then $x^4+1>1>x$. If $|x|geq 1$, then $x^4+1>x^4geq x$. In both cases
    $$
    x^4-x+1>0
    $$
    so the equation $x^4-x+1=0$ have no real solutions.


  2. Note that $x=1$ is not a solution of original equation.







share|cite|improve this answer













  1. If $|x|<1$, then $x^4+1>1>x$. If $|x|geq 1$, then $x^4+1>x^4geq x$. In both cases
    $$
    x^4-x+1>0
    $$
    so the equation $x^4-x+1=0$ have no real solutions.


  2. Note that $x=1$ is not a solution of original equation.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 26 '13 at 10:38









Norbert

45.6k773158




45.6k773158












  • Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $displaystyle tan^{-1}(infty)+tan^{-1}(1) = frac{pi}{2}+frac{pi}{4} = frac{3pi}{4}$ which is exactly equaal to $R.H.S$
    – juantheron
    Jan 26 '13 at 10:46








  • 2




    Expression $tan^{-1}(infty)$ makes no sense. Even if you are talking about limits it is not ok because $$limlimits_{xto 1+}tan^{-1}frac{x}{1-x^2}=frac{pi}{2}$$$$limlimits_{xto 1-}tan^{-1}frac{x}{1-x^2}=-frac{pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $frac{x}{1-x^2}$
    – Norbert
    Jan 26 '13 at 10:51












  • Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a)
    – juantheron
    Jan 26 '13 at 11:16




















  • Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $displaystyle tan^{-1}(infty)+tan^{-1}(1) = frac{pi}{2}+frac{pi}{4} = frac{3pi}{4}$ which is exactly equaal to $R.H.S$
    – juantheron
    Jan 26 '13 at 10:46








  • 2




    Expression $tan^{-1}(infty)$ makes no sense. Even if you are talking about limits it is not ok because $$limlimits_{xto 1+}tan^{-1}frac{x}{1-x^2}=frac{pi}{2}$$$$limlimits_{xto 1-}tan^{-1}frac{x}{1-x^2}=-frac{pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $frac{x}{1-x^2}$
    – Norbert
    Jan 26 '13 at 10:51












  • Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a)
    – juantheron
    Jan 26 '13 at 11:16


















Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $displaystyle tan^{-1}(infty)+tan^{-1}(1) = frac{pi}{2}+frac{pi}{4} = frac{3pi}{4}$ which is exactly equaal to $R.H.S$
– juantheron
Jan 26 '13 at 10:46






Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $displaystyle tan^{-1}(infty)+tan^{-1}(1) = frac{pi}{2}+frac{pi}{4} = frac{3pi}{4}$ which is exactly equaal to $R.H.S$
– juantheron
Jan 26 '13 at 10:46






2




2




Expression $tan^{-1}(infty)$ makes no sense. Even if you are talking about limits it is not ok because $$limlimits_{xto 1+}tan^{-1}frac{x}{1-x^2}=frac{pi}{2}$$$$limlimits_{xto 1-}tan^{-1}frac{x}{1-x^2}=-frac{pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $frac{x}{1-x^2}$
– Norbert
Jan 26 '13 at 10:51






Expression $tan^{-1}(infty)$ makes no sense. Even if you are talking about limits it is not ok because $$limlimits_{xto 1+}tan^{-1}frac{x}{1-x^2}=frac{pi}{2}$$$$limlimits_{xto 1-}tan^{-1}frac{x}{1-x^2}=-frac{pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $frac{x}{1-x^2}$
– Norbert
Jan 26 '13 at 10:51














Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a)
– juantheron
Jan 26 '13 at 11:16






Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a)
– juantheron
Jan 26 '13 at 11:16




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f287220%2fcalculation-of-x-in-tan-1-left-fracx1-x2-right-tan-1-left-fra%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

To store a contact into the json file from server.js file using a class in NodeJS

Image
0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
Read more

Redirect URL with Chrome Remote Debugging Android Devices

Image
1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
Read more

Dieringhausen

Image
Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
Read more