Number of solutions of the equation $cos(pisqrt{x-4})cos(pisqrt{x})=1$
$begingroup$
Find the number of solutions of the equation $cos(pisqrt{x-4})cos(pisqrt{x})=1$
begin{align}
2cos(pisqrt{x-4})&.cos(pisqrt{x})=2\impliescosBig[pi(sqrt{x-4}+sqrt{x})Big]&+cosBig[pi(sqrt{x-4}-sqrt{x})Big]=2\
impliescosBig[pi(sqrt{x-4}+sqrt{x})Big]=1quad&&quadcosBig[pi(sqrt{x-4}-sqrt{x})Big]=1\
pi(sqrt{x-4}+sqrt{x})=2npiquad&&quadpi(sqrt{x-4}-sqrt{x})=2mpi\
sqrt{x-4}+sqrt{x}=2nquad&&quadsqrt{x-4}-sqrt{x}=2m\
2sqrt{x}=2(n-m)quad&&quad2sqrt{x-4}=2(n+m)\
sqrt{x}=n-mquad&&quadsqrt{x-4}=n+mquad&quad xgeq4
end{align}
How do I properly find the solutions ?
Or can I simply say
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nmimplies nm=-1\
implies x=bigg[n+frac{1}{n}bigg]^2inmathbb{Z}implies n,frac{1}{n}inmathbb{Z}\
implies nneq0implies n=1,x=4text{ is the only solution}
$$
trigonometry
asked Dec 4 '18 at 18:07
ss1729ss1729
1,97511023
$endgroup$
add a comment |
$begingroup$
Find the number of solutions of the equation $cos(pisqrt{x-4})cos(pisqrt{x})=1$
begin{align}
2cos(pisqrt{x-4})&.cos(pisqrt{x})=2\impliescosBig[pi(sqrt{x-4}+sqrt{x})Big]&+cosBig[pi(sqrt{x-4}-sqrt{x})Big]=2\
impliescosBig[pi(sqrt{x-4}+sqrt{x})Big]=1quad&&quadcosBig[pi(sqrt{x-4}-sqrt{x})Big]=1\
pi(sqrt{x-4}+sqrt{x})=2npiquad&&quadpi(sqrt{x-4}-sqrt{x})=2mpi\
sqrt{x-4}+sqrt{x}=2nquad&&quadsqrt{x-4}-sqrt{x}=2m\
2sqrt{x}=2(n-m)quad&&quad2sqrt{x-4}=2(n+m)\
sqrt{x}=n-mquad&&quadsqrt{x-4}=n+mquad&quad xgeq4
end{align}
How do I properly find the solutions ?
Or can I simply say
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nmimplies nm=-1\
implies x=bigg[n+frac{1}{n}bigg]^2inmathbb{Z}implies n,frac{1}{n}inmathbb{Z}\
implies nneq0implies n=1,x=4text{ is the only solution}
$$
trigonometry
asked Dec 4 '18 at 18:07
ss1729ss1729
1,97511023
$endgroup$
add a comment |
$begingroup$
Find the number of solutions of the equation $cos(pisqrt{x-4})cos(pisqrt{x})=1$
begin{align}
2cos(pisqrt{x-4})&.cos(pisqrt{x})=2\impliescosBig[pi(sqrt{x-4}+sqrt{x})Big]&+cosBig[pi(sqrt{x-4}-sqrt{x})Big]=2\
impliescosBig[pi(sqrt{x-4}+sqrt{x})Big]=1quad&&quadcosBig[pi(sqrt{x-4}-sqrt{x})Big]=1\
pi(sqrt{x-4}+sqrt{x})=2npiquad&&quadpi(sqrt{x-4}-sqrt{x})=2mpi\
sqrt{x-4}+sqrt{x}=2nquad&&quadsqrt{x-4}-sqrt{x}=2m\
2sqrt{x}=2(n-m)quad&&quad2sqrt{x-4}=2(n+m)\
sqrt{x}=n-mquad&&quadsqrt{x-4}=n+mquad&quad xgeq4
end{align}
How do I properly find the solutions ?
Or can I simply say
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nmimplies nm=-1\
implies x=bigg[n+frac{1}{n}bigg]^2inmathbb{Z}implies n,frac{1}{n}inmathbb{Z}\
implies nneq0implies n=1,x=4text{ is the only solution}
$$
trigonometry
asked Dec 4 '18 at 18:07
ss1729ss1729
1,97511023
$endgroup$
Find the number of solutions of the equation $cos(pisqrt{x-4})cos(pisqrt{x})=1$
begin{align}
2cos(pisqrt{x-4})&.cos(pisqrt{x})=2\impliescosBig[pi(sqrt{x-4}+sqrt{x})Big]&+cosBig[pi(sqrt{x-4}-sqrt{x})Big]=2\
impliescosBig[pi(sqrt{x-4}+sqrt{x})Big]=1quad&&quadcosBig[pi(sqrt{x-4}-sqrt{x})Big]=1\
pi(sqrt{x-4}+sqrt{x})=2npiquad&&quadpi(sqrt{x-4}-sqrt{x})=2mpi\
sqrt{x-4}+sqrt{x}=2nquad&&quadsqrt{x-4}-sqrt{x}=2m\
2sqrt{x}=2(n-m)quad&&quad2sqrt{x-4}=2(n+m)\
sqrt{x}=n-mquad&&quadsqrt{x-4}=n+mquad&quad xgeq4
end{align}
How do I properly find the solutions ?
Or can I simply say
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nmimplies nm=-1\
implies x=bigg[n+frac{1}{n}bigg]^2inmathbb{Z}implies n,frac{1}{n}inmathbb{Z}\
implies nneq0implies n=1,x=4text{ is the only solution}
$$
trigonometry
trigonometry
asked Dec 4 '18 at 18:07
ss1729ss1729
1,97511023
asked Dec 4 '18 at 18:07
ss1729ss1729
1,97511023
asked Dec 4 '18 at 18:07
ss1729ss1729
1,97511023
asked Dec 4 '18 at 18:07
ss1729ss1729
1,97511023
asked Dec 4 '18 at 18:07
ss1729ss1729
1,97511023
1,97511023
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
More simply we need as a necessary condition
- $pisqrt{x-4}=k_1pi$
- $pisqrt{x}=k_2pi$
that is
- $sqrt{x-4}=k_1 implies x=k_1^2+4$
- $sqrt{x}=k_2implies x=k_2^2$
that is
$$k_1^2+4=k_2^2 iff k_2^2-k_1^2=4$$
and the only possible solutions is $k_1=0 implies x=4$ which works by inspection.
answered Dec 4 '18 at 18:14
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
$$(a+b)(a-b)=4$$ so, both $a+b,a-b$ are even $$1=dfrac{a+b}2cdotdfrac{a-b}2$$ $$implies dfrac{a+b}2=dfrac{a-b}2=pm1implies b=0$$
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 6:25
$begingroup$
@labbhattacharjee I've used that the distance between squares grows up as $2k+1$ and therefore we onlu need to consider $0,1,4$. That's also a nice way. Thanks, Bye
$endgroup$
– gimusi
Dec 5 '18 at 7:49
add a comment |
$begingroup$
I suppose $x$ is real in the following and that $cos$ is the function $cos:Bbb RtoBbb R$. (There is an other function $cos:Bbb CtoBbb C$, to use it i have to ask for the branch of the square root(s) first.)
The two $cos$ functions in the product (evaluated at those two places) must have (in a correlated way) the value $pm 1$. This makes things easier maybe to decide that $sqrt x$ and $sqrt {x-4}$ are two integers of same parity. In particular $xge 4$. Starting with the perfect square $2^2=4$ the distance between two perfect squares is $ge 3^2-2^2=9-4=5$, so it is at least $5$. So the bigger perfect square $x$ (among $x,x-4$) is at most $4$. We get the solution $x=2^2=4$. (There is no $x=1^2$ or $x=0^2$ as solution, because $x-4<0$.)
answered Dec 4 '18 at 18:19
dan_fuleadan_fulea
6,6431312
$endgroup$
add a comment |
$begingroup$
$cos(pisqrt{x-4})cos(pisqrt x) = 1$
implies
$cos(pisqrt{x-4}) = cos(pisqrt x) = 1$
or
$cos(pisqrt{x-4}) = cos(pisqrt x) = -1$
in either case
$sqrt{x-4}, sqrt x$ are both integers. In fact, they are both even, or they are both odd. But that isn't entirely relevant.
There is only one pair of integers that are perfect squares and separated by a distance of 4.
$x = 4$
answered Dec 4 '18 at 18:19
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
Maybe you mean $x=4$? The expression in the OP is different.
$endgroup$
– gimusi
Dec 4 '18 at 18:20
add a comment |
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3 Answers
3
active
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3 Answers
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active
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$begingroup$
More simply we need as a necessary condition
- $pisqrt{x-4}=k_1pi$
- $pisqrt{x}=k_2pi$
that is
- $sqrt{x-4}=k_1 implies x=k_1^2+4$
- $sqrt{x}=k_2implies x=k_2^2$
that is
$$k_1^2+4=k_2^2 iff k_2^2-k_1^2=4$$
and the only possible solutions is $k_1=0 implies x=4$ which works by inspection.
answered Dec 4 '18 at 18:14
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
$$(a+b)(a-b)=4$$ so, both $a+b,a-b$ are even $$1=dfrac{a+b}2cdotdfrac{a-b}2$$ $$implies dfrac{a+b}2=dfrac{a-b}2=pm1implies b=0$$
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 6:25
$begingroup$
@labbhattacharjee I've used that the distance between squares grows up as $2k+1$ and therefore we onlu need to consider $0,1,4$. That's also a nice way. Thanks, Bye
$endgroup$
– gimusi
Dec 5 '18 at 7:49
add a comment |
$begingroup$
More simply we need as a necessary condition
- $pisqrt{x-4}=k_1pi$
- $pisqrt{x}=k_2pi$
that is
- $sqrt{x-4}=k_1 implies x=k_1^2+4$
- $sqrt{x}=k_2implies x=k_2^2$
that is
$$k_1^2+4=k_2^2 iff k_2^2-k_1^2=4$$
and the only possible solutions is $k_1=0 implies x=4$ which works by inspection.
answered Dec 4 '18 at 18:14
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
$$(a+b)(a-b)=4$$ so, both $a+b,a-b$ are even $$1=dfrac{a+b}2cdotdfrac{a-b}2$$ $$implies dfrac{a+b}2=dfrac{a-b}2=pm1implies b=0$$
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 6:25
$begingroup$
@labbhattacharjee I've used that the distance between squares grows up as $2k+1$ and therefore we onlu need to consider $0,1,4$. That's also a nice way. Thanks, Bye
$endgroup$
– gimusi
Dec 5 '18 at 7:49
add a comment |
$begingroup$
More simply we need as a necessary condition
- $pisqrt{x-4}=k_1pi$
- $pisqrt{x}=k_2pi$
that is
- $sqrt{x-4}=k_1 implies x=k_1^2+4$
- $sqrt{x}=k_2implies x=k_2^2$
that is
$$k_1^2+4=k_2^2 iff k_2^2-k_1^2=4$$
and the only possible solutions is $k_1=0 implies x=4$ which works by inspection.
answered Dec 4 '18 at 18:14
gimusigimusi
92.9k84494
$endgroup$
More simply we need as a necessary condition
- $pisqrt{x-4}=k_1pi$
- $pisqrt{x}=k_2pi$
that is
- $sqrt{x-4}=k_1 implies x=k_1^2+4$
- $sqrt{x}=k_2implies x=k_2^2$
that is
$$k_1^2+4=k_2^2 iff k_2^2-k_1^2=4$$
and the only possible solutions is $k_1=0 implies x=4$ which works by inspection.
answered Dec 4 '18 at 18:14
gimusigimusi
92.9k84494
answered Dec 4 '18 at 18:14
gimusigimusi
92.9k84494
answered Dec 4 '18 at 18:14
gimusigimusi
92.9k84494
answered Dec 4 '18 at 18:14
gimusigimusi
92.9k84494
92.9k84494
$begingroup$
$$(a+b)(a-b)=4$$ so, both $a+b,a-b$ are even $$1=dfrac{a+b}2cdotdfrac{a-b}2$$ $$implies dfrac{a+b}2=dfrac{a-b}2=pm1implies b=0$$
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 6:25
$begingroup$
@labbhattacharjee I've used that the distance between squares grows up as $2k+1$ and therefore we onlu need to consider $0,1,4$. That's also a nice way. Thanks, Bye
$endgroup$
– gimusi
Dec 5 '18 at 7:49
add a comment |
$begingroup$
$$(a+b)(a-b)=4$$ so, both $a+b,a-b$ are even $$1=dfrac{a+b}2cdotdfrac{a-b}2$$ $$implies dfrac{a+b}2=dfrac{a-b}2=pm1implies b=0$$
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 6:25
$begingroup$
@labbhattacharjee I've used that the distance between squares grows up as $2k+1$ and therefore we onlu need to consider $0,1,4$. That's also a nice way. Thanks, Bye
$endgroup$
– gimusi
Dec 5 '18 at 7:49
$begingroup$
$$(a+b)(a-b)=4$$ so, both $a+b,a-b$ are even $$1=dfrac{a+b}2cdotdfrac{a-b}2$$ $$implies dfrac{a+b}2=dfrac{a-b}2=pm1implies b=0$$
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 6:25
$begingroup$
$$(a+b)(a-b)=4$$ so, both $a+b,a-b$ are even $$1=dfrac{a+b}2cdotdfrac{a-b}2$$ $$implies dfrac{a+b}2=dfrac{a-b}2=pm1implies b=0$$
$endgroup$
– lab bhattacharjee
Dec 5 '18 at 6:25
$begingroup$
@labbhattacharjee I've used that the distance between squares grows up as $2k+1$ and therefore we onlu need to consider $0,1,4$. That's also a nice way. Thanks, Bye
$endgroup$
– gimusi
Dec 5 '18 at 7:49
$begingroup$
@labbhattacharjee I've used that the distance between squares grows up as $2k+1$ and therefore we onlu need to consider $0,1,4$. That's also a nice way. Thanks, Bye
$endgroup$
– gimusi
Dec 5 '18 at 7:49
add a comment |
$begingroup$
I suppose $x$ is real in the following and that $cos$ is the function $cos:Bbb RtoBbb R$. (There is an other function $cos:Bbb CtoBbb C$, to use it i have to ask for the branch of the square root(s) first.)
The two $cos$ functions in the product (evaluated at those two places) must have (in a correlated way) the value $pm 1$. This makes things easier maybe to decide that $sqrt x$ and $sqrt {x-4}$ are two integers of same parity. In particular $xge 4$. Starting with the perfect square $2^2=4$ the distance between two perfect squares is $ge 3^2-2^2=9-4=5$, so it is at least $5$. So the bigger perfect square $x$ (among $x,x-4$) is at most $4$. We get the solution $x=2^2=4$. (There is no $x=1^2$ or $x=0^2$ as solution, because $x-4<0$.)
answered Dec 4 '18 at 18:19
dan_fuleadan_fulea
6,6431312
$endgroup$
add a comment |
$begingroup$
I suppose $x$ is real in the following and that $cos$ is the function $cos:Bbb RtoBbb R$. (There is an other function $cos:Bbb CtoBbb C$, to use it i have to ask for the branch of the square root(s) first.)
The two $cos$ functions in the product (evaluated at those two places) must have (in a correlated way) the value $pm 1$. This makes things easier maybe to decide that $sqrt x$ and $sqrt {x-4}$ are two integers of same parity. In particular $xge 4$. Starting with the perfect square $2^2=4$ the distance between two perfect squares is $ge 3^2-2^2=9-4=5$, so it is at least $5$. So the bigger perfect square $x$ (among $x,x-4$) is at most $4$. We get the solution $x=2^2=4$. (There is no $x=1^2$ or $x=0^2$ as solution, because $x-4<0$.)
answered Dec 4 '18 at 18:19
dan_fuleadan_fulea
6,6431312
$endgroup$
add a comment |
$begingroup$
I suppose $x$ is real in the following and that $cos$ is the function $cos:Bbb RtoBbb R$. (There is an other function $cos:Bbb CtoBbb C$, to use it i have to ask for the branch of the square root(s) first.)
The two $cos$ functions in the product (evaluated at those two places) must have (in a correlated way) the value $pm 1$. This makes things easier maybe to decide that $sqrt x$ and $sqrt {x-4}$ are two integers of same parity. In particular $xge 4$. Starting with the perfect square $2^2=4$ the distance between two perfect squares is $ge 3^2-2^2=9-4=5$, so it is at least $5$. So the bigger perfect square $x$ (among $x,x-4$) is at most $4$. We get the solution $x=2^2=4$. (There is no $x=1^2$ or $x=0^2$ as solution, because $x-4<0$.)
answered Dec 4 '18 at 18:19
dan_fuleadan_fulea
6,6431312
$endgroup$
I suppose $x$ is real in the following and that $cos$ is the function $cos:Bbb RtoBbb R$. (There is an other function $cos:Bbb CtoBbb C$, to use it i have to ask for the branch of the square root(s) first.)
The two $cos$ functions in the product (evaluated at those two places) must have (in a correlated way) the value $pm 1$. This makes things easier maybe to decide that $sqrt x$ and $sqrt {x-4}$ are two integers of same parity. In particular $xge 4$. Starting with the perfect square $2^2=4$ the distance between two perfect squares is $ge 3^2-2^2=9-4=5$, so it is at least $5$. So the bigger perfect square $x$ (among $x,x-4$) is at most $4$. We get the solution $x=2^2=4$. (There is no $x=1^2$ or $x=0^2$ as solution, because $x-4<0$.)
answered Dec 4 '18 at 18:19
dan_fuleadan_fulea
6,6431312
answered Dec 4 '18 at 18:19
dan_fuleadan_fulea
6,6431312
answered Dec 4 '18 at 18:19
dan_fuleadan_fulea
6,6431312
answered Dec 4 '18 at 18:19
dan_fuleadan_fulea
6,6431312
6,6431312
add a comment |
add a comment |
$begingroup$
$cos(pisqrt{x-4})cos(pisqrt x) = 1$
implies
$cos(pisqrt{x-4}) = cos(pisqrt x) = 1$
or
$cos(pisqrt{x-4}) = cos(pisqrt x) = -1$
in either case
$sqrt{x-4}, sqrt x$ are both integers. In fact, they are both even, or they are both odd. But that isn't entirely relevant.
There is only one pair of integers that are perfect squares and separated by a distance of 4.
$x = 4$
answered Dec 4 '18 at 18:19
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
Maybe you mean $x=4$? The expression in the OP is different.
$endgroup$
– gimusi
Dec 4 '18 at 18:20
add a comment |
$begingroup$
$cos(pisqrt{x-4})cos(pisqrt x) = 1$
implies
$cos(pisqrt{x-4}) = cos(pisqrt x) = 1$
or
$cos(pisqrt{x-4}) = cos(pisqrt x) = -1$
in either case
$sqrt{x-4}, sqrt x$ are both integers. In fact, they are both even, or they are both odd. But that isn't entirely relevant.
There is only one pair of integers that are perfect squares and separated by a distance of 4.
$x = 4$
answered Dec 4 '18 at 18:19
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
Maybe you mean $x=4$? The expression in the OP is different.
$endgroup$
– gimusi
Dec 4 '18 at 18:20
add a comment |
$begingroup$
$cos(pisqrt{x-4})cos(pisqrt x) = 1$
implies
$cos(pisqrt{x-4}) = cos(pisqrt x) = 1$
or
$cos(pisqrt{x-4}) = cos(pisqrt x) = -1$
in either case
$sqrt{x-4}, sqrt x$ are both integers. In fact, they are both even, or they are both odd. But that isn't entirely relevant.
There is only one pair of integers that are perfect squares and separated by a distance of 4.
$x = 4$
answered Dec 4 '18 at 18:19
Doug MDoug M
45.3k31954
$endgroup$
$cos(pisqrt{x-4})cos(pisqrt x) = 1$
implies
$cos(pisqrt{x-4}) = cos(pisqrt x) = 1$
or
$cos(pisqrt{x-4}) = cos(pisqrt x) = -1$
in either case
$sqrt{x-4}, sqrt x$ are both integers. In fact, they are both even, or they are both odd. But that isn't entirely relevant.
There is only one pair of integers that are perfect squares and separated by a distance of 4.
$x = 4$
answered Dec 4 '18 at 18:19
Doug MDoug M
45.3k31954
edited Dec 4 '18 at 18:41
answered Dec 4 '18 at 18:19
Doug MDoug M
45.3k31954
answered Dec 4 '18 at 18:19
Doug MDoug M
45.3k31954
answered Dec 4 '18 at 18:19
Doug MDoug M
45.3k31954
45.3k31954
$begingroup$
Maybe you mean $x=4$? The expression in the OP is different.
$endgroup$
– gimusi
Dec 4 '18 at 18:20
add a comment |
$begingroup$
Maybe you mean $x=4$? The expression in the OP is different.
$endgroup$
– gimusi
Dec 4 '18 at 18:20
$begingroup$
Maybe you mean $x=4$? The expression in the OP is different.
$endgroup$
– gimusi
Dec 4 '18 at 18:20
$begingroup$
Maybe you mean $x=4$? The expression in the OP is different.
$endgroup$
– gimusi
Dec 4 '18 at 18:20
add a comment |
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