Find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.

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Let $f(x)=x^2$ and $g(x)=sin(x) forall xin mathbb R$.
Then find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.




Solution. From the given data, we have
$(fcirc gcirc gcirc f)(x)=(sin(sin(x^2)))^2$ and $(gcirc gcirc f)(x)=sin(sin(x^2))$.



Now we have to find the values of $x$ satisfying
$$(sin(sin(x^2)))^2=sin(sin(x^2))tag1$$



Let $sin(sin(x^2))=t$, then $(1)$ becomes $t^2=timplies t=0 text or 1$



Case 1: $sin(sin(x^2))=0implies sin(x^2)=npi, nin mathbb Z$. Need help in proceeding further



Case 2: $sin(sin(x^2))=1implies sin(x^2)=npi+(-1)^n left (fracpi2 right)$. Need help in proceeding further




trigonometry systems-of-equations roots




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  • Please edit the text in RED"Need help in proceeding further"
    – PK Styles
    13 hours ago










  • do you mean $f(g(g(f(x))))=g(g(f(x)))$
    – Henry Lee
    13 hours ago










  • @HenryLee:Yeah,you got it right
    – PK Styles
    13 hours ago






  • 1




    Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
    – Cameron Buie
    13 hours ago














up vote
2
down vote

favorite













Let $f(x)=x^2$ and $g(x)=sin(x) forall xin mathbb R$.
Then find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.




Solution. From the given data, we have
$(fcirc gcirc gcirc f)(x)=(sin(sin(x^2)))^2$ and $(gcirc gcirc f)(x)=sin(sin(x^2))$.



Now we have to find the values of $x$ satisfying
$$(sin(sin(x^2)))^2=sin(sin(x^2))tag1$$



Let $sin(sin(x^2))=t$, then $(1)$ becomes $t^2=timplies t=0 text or 1$



Case 1: $sin(sin(x^2))=0implies sin(x^2)=npi, nin mathbb Z$. Need help in proceeding further



Case 2: $sin(sin(x^2))=1implies sin(x^2)=npi+(-1)^n left (fracpi2 right)$. Need help in proceeding further




trigonometry systems-of-equations roots




share|cite|improve this question





















  • Please edit the text in RED"Need help in proceeding further"
    – PK Styles
    13 hours ago










  • do you mean $f(g(g(f(x))))=g(g(f(x)))$
    – Henry Lee
    13 hours ago










  • @HenryLee:Yeah,you got it right
    – PK Styles
    13 hours ago






  • 1




    Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
    – Cameron Buie
    13 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Let $f(x)=x^2$ and $g(x)=sin(x) forall xin mathbb R$.
Then find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.




Solution. From the given data, we have
$(fcirc gcirc gcirc f)(x)=(sin(sin(x^2)))^2$ and $(gcirc gcirc f)(x)=sin(sin(x^2))$.



Now we have to find the values of $x$ satisfying
$$(sin(sin(x^2)))^2=sin(sin(x^2))tag1$$



Let $sin(sin(x^2))=t$, then $(1)$ becomes $t^2=timplies t=0 text or 1$



Case 1: $sin(sin(x^2))=0implies sin(x^2)=npi, nin mathbb Z$. Need help in proceeding further



Case 2: $sin(sin(x^2))=1implies sin(x^2)=npi+(-1)^n left (fracpi2 right)$. Need help in proceeding further




trigonometry systems-of-equations roots




share|cite|improve this question














Let $f(x)=x^2$ and $g(x)=sin(x) forall xin mathbb R$.
Then find the value of all $x$ satisfying $(fcirc gcirc gcirc f)(x)=(gcirc gcirc f)(x)$, where $(fcirc g)(x)=f(g(x))$.




Solution. From the given data, we have
$(fcirc gcirc gcirc f)(x)=(sin(sin(x^2)))^2$ and $(gcirc gcirc f)(x)=sin(sin(x^2))$.



Now we have to find the values of $x$ satisfying
$$(sin(sin(x^2)))^2=sin(sin(x^2))tag1$$



Let $sin(sin(x^2))=t$, then $(1)$ becomes $t^2=timplies t=0 text or 1$



Case 1: $sin(sin(x^2))=0implies sin(x^2)=npi, nin mathbb Z$. Need help in proceeding further



Case 2: $sin(sin(x^2))=1implies sin(x^2)=npi+(-1)^n left (fracpi2 right)$. Need help in proceeding further





trigonometry systems-of-equations roots





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 13 hours ago









Rócherz

2,1811417




2,1811417









asked 13 hours ago









PK Styles

1,392524




1,392524











  • Please edit the text in RED"Need help in proceeding further"
    – PK Styles
    13 hours ago










  • do you mean $f(g(g(f(x))))=g(g(f(x)))$
    – Henry Lee
    13 hours ago










  • @HenryLee:Yeah,you got it right
    – PK Styles
    13 hours ago






  • 1




    Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
    – Cameron Buie
    13 hours ago
















  • Please edit the text in RED"Need help in proceeding further"
    – PK Styles
    13 hours ago










  • do you mean $f(g(g(f(x))))=g(g(f(x)))$
    – Henry Lee
    13 hours ago










  • @HenryLee:Yeah,you got it right
    – PK Styles
    13 hours ago






  • 1




    Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
    – Cameron Buie
    13 hours ago















Please edit the text in RED"Need help in proceeding further"
– PK Styles
13 hours ago




Please edit the text in RED"Need help in proceeding further"
– PK Styles
13 hours ago












do you mean $f(g(g(f(x))))=g(g(f(x)))$
– Henry Lee
13 hours ago




do you mean $f(g(g(f(x))))=g(g(f(x)))$
– Henry Lee
13 hours ago












@HenryLee:Yeah,you got it right
– PK Styles
13 hours ago




@HenryLee:Yeah,you got it right
– PK Styles
13 hours ago




1




1




Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
– Cameron Buie
13 hours ago




Your solution for Case 2, while accurate, is not optimal. For example, the solution is the same for $n=0$ and $n=1.$ Ideally, we'd like a different solution for each $n,$ such as by $fracpi2+2npi$ for $ninBbb Z.$
– Cameron Buie
13 hours ago










1 Answer
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up vote
3
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Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.



Can you solve case $2$ in a similar manner?



EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.






share|cite|improve this answer























  • :Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
    – PK Styles
    12 hours ago










  • Exactly ! Note that necessarily, $n$ is non-negative.
    – Suzet
    12 hours ago










  • :will you please clarify case 2 also?
    – PK Styles
    12 hours ago










  • @PKStyles I added the case $2$ in my answer.
    – Suzet
    3 hours ago










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active

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up vote
3
down vote













Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.



Can you solve case $2$ in a similar manner?



EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.






share|cite|improve this answer























  • :Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
    – PK Styles
    12 hours ago










  • Exactly ! Note that necessarily, $n$ is non-negative.
    – Suzet
    12 hours ago










  • :will you please clarify case 2 also?
    – PK Styles
    12 hours ago










  • @PKStyles I added the case $2$ in my answer.
    – Suzet
    3 hours ago














up vote
3
down vote













Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.



Can you solve case $2$ in a similar manner?



EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.






share|cite|improve this answer























  • :Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
    – PK Styles
    12 hours ago










  • Exactly ! Note that necessarily, $n$ is non-negative.
    – Suzet
    12 hours ago










  • :will you please clarify case 2 also?
    – PK Styles
    12 hours ago










  • @PKStyles I added the case $2$ in my answer.
    – Suzet
    3 hours ago












up vote
3
down vote










up vote
3
down vote









Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.



Can you solve case $2$ in a similar manner?



EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.






share|cite|improve this answer















Recall that $sin$ is a bounded function between $-1$ and $1$. So the only way that $sin(x^2)$ is of the form $npi$ for some $nin mathbbZ$ is when $n=0$. You can then solve this easily for $x$.



Can you solve case $2$ in a similar manner?



EDIT: As for case $2$, as @CameronBuie remarked in the comment, the usual writing of the solutions of $sin(z)=0$ is $z=fracpi2+2npi$. Note that no real number of this form lie between $1$ and $-1$ (the least such positive real number is $fracpi2=1.57ldots$ and the greatest negative real number of this form is $fracpi2-2pi=-4.71ldots$). Hence, there is no solution in the case $2$ where $sin(x^2)=fracpi2+2npi$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago


























answered 13 hours ago









Suzet

2,003324




2,003324











  • :Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
    – PK Styles
    12 hours ago










  • Exactly ! Note that necessarily, $n$ is non-negative.
    – Suzet
    12 hours ago










  • :will you please clarify case 2 also?
    – PK Styles
    12 hours ago










  • @PKStyles I added the case $2$ in my answer.
    – Suzet
    3 hours ago
















  • :Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
    – PK Styles
    12 hours ago










  • Exactly ! Note that necessarily, $n$ is non-negative.
    – Suzet
    12 hours ago










  • :will you please clarify case 2 also?
    – PK Styles
    12 hours ago










  • @PKStyles I added the case $2$ in my answer.
    – Suzet
    3 hours ago















:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago




:Following your comment ,$sin(x^2)=0implies$ $x^2=npi implies x=pm sqrt npi$.Am i correct?
– PK Styles
12 hours ago












Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago




Exactly ! Note that necessarily, $n$ is non-negative.
– Suzet
12 hours ago












:will you please clarify case 2 also?
– PK Styles
12 hours ago




:will you please clarify case 2 also?
– PK Styles
12 hours ago












@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago




@PKStyles I added the case $2$ in my answer.
– Suzet
3 hours ago












 

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