Values of $x$ satisfying $sin xcdotcos^3 x>sin^3xcdotcos x$
$begingroup$
For what values of $x$ between $0$ and $pi$ does the inequality $sin xcdotcos^3 x>sin^3xcdotcos x$ hold?
My Attempt
$$
sin xcos xcdot(cos^2x-sin^2x)=frac{1}{2}cdotsin2xcdotcos2x=frac{1}{4}cdotsin4x>0impliessin4x>0\
xin(0,pi)implies4xin(0,4pi)\
4xin(0,pi)cup(2pi,3pi)implies xinBig(0,frac{pi}{4}Big)cupBig(frac{pi}{2},frac{3pi}{4}Big)
$$
But, my reference gives the solution, $xinBig(0,dfrac{pi}{4}Big)cupBig(dfrac{3pi}{4},piBig)$, where am I going wrong with my attempt?
trigonometry inequality
edited Dec 1 '18 at 19:10
TheSimpliFire
12.6k62360
asked Dec 1 '18 at 18:56
ss1729ss1729
1,8541723
$endgroup$
add a comment |
$begingroup$
For what values of $x$ between $0$ and $pi$ does the inequality $sin xcdotcos^3 x>sin^3xcdotcos x$ hold?
My Attempt
$$
sin xcos xcdot(cos^2x-sin^2x)=frac{1}{2}cdotsin2xcdotcos2x=frac{1}{4}cdotsin4x>0impliessin4x>0\
xin(0,pi)implies4xin(0,4pi)\
4xin(0,pi)cup(2pi,3pi)implies xinBig(0,frac{pi}{4}Big)cupBig(frac{pi}{2},frac{3pi}{4}Big)
$$
But, my reference gives the solution, $xinBig(0,dfrac{pi}{4}Big)cupBig(dfrac{3pi}{4},piBig)$, where am I going wrong with my attempt?
trigonometry inequality
edited Dec 1 '18 at 19:10
TheSimpliFire
12.6k62360
asked Dec 1 '18 at 18:56
ss1729ss1729
1,8541723
$endgroup$
add a comment |
$begingroup$
For what values of $x$ between $0$ and $pi$ does the inequality $sin xcdotcos^3 x>sin^3xcdotcos x$ hold?
My Attempt
$$
sin xcos xcdot(cos^2x-sin^2x)=frac{1}{2}cdotsin2xcdotcos2x=frac{1}{4}cdotsin4x>0impliessin4x>0\
xin(0,pi)implies4xin(0,4pi)\
4xin(0,pi)cup(2pi,3pi)implies xinBig(0,frac{pi}{4}Big)cupBig(frac{pi}{2},frac{3pi}{4}Big)
$$
But, my reference gives the solution, $xinBig(0,dfrac{pi}{4}Big)cupBig(dfrac{3pi}{4},piBig)$, where am I going wrong with my attempt?
trigonometry inequality
edited Dec 1 '18 at 19:10
TheSimpliFire
12.6k62360
asked Dec 1 '18 at 18:56
ss1729ss1729
1,8541723
$endgroup$
For what values of $x$ between $0$ and $pi$ does the inequality $sin xcdotcos^3 x>sin^3xcdotcos x$ hold?
My Attempt
$$
sin xcos xcdot(cos^2x-sin^2x)=frac{1}{2}cdotsin2xcdotcos2x=frac{1}{4}cdotsin4x>0impliessin4x>0\
xin(0,pi)implies4xin(0,4pi)\
4xin(0,pi)cup(2pi,3pi)implies xinBig(0,frac{pi}{4}Big)cupBig(frac{pi}{2},frac{3pi}{4}Big)
$$
But, my reference gives the solution, $xinBig(0,dfrac{pi}{4}Big)cupBig(dfrac{3pi}{4},piBig)$, where am I going wrong with my attempt?
trigonometry inequality
trigonometry inequality
edited Dec 1 '18 at 19:10
TheSimpliFire
12.6k62360
asked Dec 1 '18 at 18:56
ss1729ss1729
1,8541723
edited Dec 1 '18 at 19:10
TheSimpliFire
12.6k62360
asked Dec 1 '18 at 18:56
ss1729ss1729
1,8541723
edited Dec 1 '18 at 19:10
TheSimpliFire
12.6k62360
edited Dec 1 '18 at 19:10
TheSimpliFire
12.6k62360
edited Dec 1 '18 at 19:10
TheSimpliFire
12.6k62360
12.6k62360
asked Dec 1 '18 at 18:56
ss1729ss1729
1,8541723
asked Dec 1 '18 at 18:56
ss1729ss1729
1,8541723
asked Dec 1 '18 at 18:56
ss1729ss1729
1,8541723
1,8541723
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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The given solution is wrong; you are correct. At $x=frac{7pi}8inleft(frac{3pi}4,piright)$, we have that $$frac14sin4x=frac14sinfrac{7pi}2=-frac14<0$$ which is a contradiction.
answered Dec 1 '18 at 19:00
TheSimpliFireTheSimpliFire
12.6k62360
$endgroup$
add a comment |
$begingroup$
Finishing what you did
$$sin(4x)>0iff$$
$$2kpi <4x<(2k+1)pi iff$$
$$frac{kpi}{2}<x<frac{kpi}{2}+frac{pi}{4}$$
$k=0$ gives $0<x<frac{pi}{4}$
and
$k=1$ gives $frac{pi}{2}<x<frac{3pi}{4}$.
answered Dec 1 '18 at 19:12
hamam_Abdallahhamam_Abdallah
38k21634
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add a comment |
$begingroup$
As an alternative for a full solution we can consider two cases
$sin x cos x >0$ that is $xin(0,pi/2)cup(pi,3pi/2)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x>sin^2x iff2sin^2 x<1$$
$$-frac{sqrt 2}2<sin x<0 ,land, 0<sin x<frac{sqrt 2}2 iff color{red}{xin(0,pi/4)}cup(pi,5pi/4)$$
$sin x cos x <0$ that is $xin(pi/2,pi)cup(3pi/2,2pi)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x<sin^2x iff2sin^2 x>1$$
$$-1<sin x<-frac{sqrt 2}2,land, frac{sqrt 2}2<sin x <1 iff color{red}{xin(pi/2,3pi/4)}cup(3pi/2,7pi/4)$$
and then your solution is correct.
answered Dec 1 '18 at 20:02
gimusigimusi
1
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The given solution is wrong; you are correct. At $x=frac{7pi}8inleft(frac{3pi}4,piright)$, we have that $$frac14sin4x=frac14sinfrac{7pi}2=-frac14<0$$ which is a contradiction.
answered Dec 1 '18 at 19:00
TheSimpliFireTheSimpliFire
12.6k62360
$endgroup$
add a comment |
$begingroup$
The given solution is wrong; you are correct. At $x=frac{7pi}8inleft(frac{3pi}4,piright)$, we have that $$frac14sin4x=frac14sinfrac{7pi}2=-frac14<0$$ which is a contradiction.
answered Dec 1 '18 at 19:00
TheSimpliFireTheSimpliFire
12.6k62360
$endgroup$
add a comment |
$begingroup$
The given solution is wrong; you are correct. At $x=frac{7pi}8inleft(frac{3pi}4,piright)$, we have that $$frac14sin4x=frac14sinfrac{7pi}2=-frac14<0$$ which is a contradiction.
answered Dec 1 '18 at 19:00
TheSimpliFireTheSimpliFire
12.6k62360
$endgroup$
The given solution is wrong; you are correct. At $x=frac{7pi}8inleft(frac{3pi}4,piright)$, we have that $$frac14sin4x=frac14sinfrac{7pi}2=-frac14<0$$ which is a contradiction.
answered Dec 1 '18 at 19:00
TheSimpliFireTheSimpliFire
12.6k62360
answered Dec 1 '18 at 19:00
TheSimpliFireTheSimpliFire
12.6k62360
answered Dec 1 '18 at 19:00
TheSimpliFireTheSimpliFire
12.6k62360
answered Dec 1 '18 at 19:00
TheSimpliFireTheSimpliFire
12.6k62360
12.6k62360
add a comment |
add a comment |
$begingroup$
Finishing what you did
$$sin(4x)>0iff$$
$$2kpi <4x<(2k+1)pi iff$$
$$frac{kpi}{2}<x<frac{kpi}{2}+frac{pi}{4}$$
$k=0$ gives $0<x<frac{pi}{4}$
and
$k=1$ gives $frac{pi}{2}<x<frac{3pi}{4}$.
answered Dec 1 '18 at 19:12
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
Finishing what you did
$$sin(4x)>0iff$$
$$2kpi <4x<(2k+1)pi iff$$
$$frac{kpi}{2}<x<frac{kpi}{2}+frac{pi}{4}$$
$k=0$ gives $0<x<frac{pi}{4}$
and
$k=1$ gives $frac{pi}{2}<x<frac{3pi}{4}$.
answered Dec 1 '18 at 19:12
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
Finishing what you did
$$sin(4x)>0iff$$
$$2kpi <4x<(2k+1)pi iff$$
$$frac{kpi}{2}<x<frac{kpi}{2}+frac{pi}{4}$$
$k=0$ gives $0<x<frac{pi}{4}$
and
$k=1$ gives $frac{pi}{2}<x<frac{3pi}{4}$.
answered Dec 1 '18 at 19:12
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
Finishing what you did
$$sin(4x)>0iff$$
$$2kpi <4x<(2k+1)pi iff$$
$$frac{kpi}{2}<x<frac{kpi}{2}+frac{pi}{4}$$
$k=0$ gives $0<x<frac{pi}{4}$
and
$k=1$ gives $frac{pi}{2}<x<frac{3pi}{4}$.
answered Dec 1 '18 at 19:12
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 1 '18 at 19:12
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 1 '18 at 19:12
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 1 '18 at 19:12
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
$begingroup$
As an alternative for a full solution we can consider two cases
$sin x cos x >0$ that is $xin(0,pi/2)cup(pi,3pi/2)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x>sin^2x iff2sin^2 x<1$$
$$-frac{sqrt 2}2<sin x<0 ,land, 0<sin x<frac{sqrt 2}2 iff color{red}{xin(0,pi/4)}cup(pi,5pi/4)$$
$sin x cos x <0$ that is $xin(pi/2,pi)cup(3pi/2,2pi)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x<sin^2x iff2sin^2 x>1$$
$$-1<sin x<-frac{sqrt 2}2,land, frac{sqrt 2}2<sin x <1 iff color{red}{xin(pi/2,3pi/4)}cup(3pi/2,7pi/4)$$
and then your solution is correct.
answered Dec 1 '18 at 20:02
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
As an alternative for a full solution we can consider two cases
$sin x cos x >0$ that is $xin(0,pi/2)cup(pi,3pi/2)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x>sin^2x iff2sin^2 x<1$$
$$-frac{sqrt 2}2<sin x<0 ,land, 0<sin x<frac{sqrt 2}2 iff color{red}{xin(0,pi/4)}cup(pi,5pi/4)$$
$sin x cos x <0$ that is $xin(pi/2,pi)cup(3pi/2,2pi)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x<sin^2x iff2sin^2 x>1$$
$$-1<sin x<-frac{sqrt 2}2,land, frac{sqrt 2}2<sin x <1 iff color{red}{xin(pi/2,3pi/4)}cup(3pi/2,7pi/4)$$
and then your solution is correct.
answered Dec 1 '18 at 20:02
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
As an alternative for a full solution we can consider two cases
$sin x cos x >0$ that is $xin(0,pi/2)cup(pi,3pi/2)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x>sin^2x iff2sin^2 x<1$$
$$-frac{sqrt 2}2<sin x<0 ,land, 0<sin x<frac{sqrt 2}2 iff color{red}{xin(0,pi/4)}cup(pi,5pi/4)$$
$sin x cos x <0$ that is $xin(pi/2,pi)cup(3pi/2,2pi)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x<sin^2x iff2sin^2 x>1$$
$$-1<sin x<-frac{sqrt 2}2,land, frac{sqrt 2}2<sin x <1 iff color{red}{xin(pi/2,3pi/4)}cup(3pi/2,7pi/4)$$
and then your solution is correct.
answered Dec 1 '18 at 20:02
gimusigimusi
1
$endgroup$
As an alternative for a full solution we can consider two cases
$sin x cos x >0$ that is $xin(0,pi/2)cup(pi,3pi/2)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x>sin^2x iff2sin^2 x<1$$
$$-frac{sqrt 2}2<sin x<0 ,land, 0<sin x<frac{sqrt 2}2 iff color{red}{xin(0,pi/4)}cup(pi,5pi/4)$$
$sin x cos x <0$ that is $xin(pi/2,pi)cup(3pi/2,2pi)$
$$sin xcdotcos^3 x>sin^3xcdotcos x iffcos^2x<sin^2x iff2sin^2 x>1$$
$$-1<sin x<-frac{sqrt 2}2,land, frac{sqrt 2}2<sin x <1 iff color{red}{xin(pi/2,3pi/4)}cup(3pi/2,7pi/4)$$
and then your solution is correct.
answered Dec 1 '18 at 20:02
gimusigimusi
1
answered Dec 1 '18 at 20:02
gimusigimusi
1
answered Dec 1 '18 at 20:02
gimusigimusi
1
answered Dec 1 '18 at 20:02
gimusigimusi
1
1
add a comment |
add a comment |
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