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













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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













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1 Answer
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1 Answer
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active

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active

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active

oldest

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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




















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