Mixing
If $log_{0.5}sin x=1-log_{0.5}cos x$ , then the number of solutions in the interval [$-2pi, 2pi$] is?
$begingroup$
In the given solution,the answer gives only two solutions. However, when a graph of sin 2x is plotted, we see that it attains a value of 1 at four points in the given interval. What am I missing? Thank you!
trigonometry logarithms
asked Jan 31 at 18:56
Darshini PoolaDarshini Poola
126
$endgroup$
|
show 6 more comments
$begingroup$
In the given solution,the answer gives only two solutions. However, when a graph of sin 2x is plotted, we see that it attains a value of 1 at four points in the given interval. What am I missing? Thank you!
trigonometry logarithms
asked Jan 31 at 18:56
Darshini PoolaDarshini Poola
126
$endgroup$
$begingroup$
i get $$-frac{7}{4}pi,-frac{3}{4}pi,frac{pi}{4},frac{5}{4}pi$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:03
$begingroup$
So do I. I wanted to know whether I was making an error.
$endgroup$
– Darshini Poola
Jan 31 at 19:05
$begingroup$
You have no error in your calculations.
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:06
$begingroup$
So is the given solution incorrect?
$endgroup$
– Darshini Poola
Jan 31 at 19:07
$begingroup$
@Dr.SonnhardGraubner: If we assume to work in $mathbb R$, $log(cdot)$ is defined only when the argument is greater than $0$, which means that both sine and cosine must be postive, in this equation. A valid solution must be in the first quadrant. Correct?
$endgroup$
– Matteo
Jan 31 at 19:09
|
show 6 more comments
$begingroup$
In the given solution,the answer gives only two solutions. However, when a graph of sin 2x is plotted, we see that it attains a value of 1 at four points in the given interval. What am I missing? Thank you!
trigonometry logarithms
asked Jan 31 at 18:56
Darshini PoolaDarshini Poola
126
$endgroup$
In the given solution,the answer gives only two solutions. However, when a graph of sin 2x is plotted, we see that it attains a value of 1 at four points in the given interval. What am I missing? Thank you!
trigonometry logarithms
trigonometry logarithms
asked Jan 31 at 18:56
Darshini PoolaDarshini Poola
126
asked Jan 31 at 18:56
Darshini PoolaDarshini Poola
126
edited Feb 1 at 19:21
Darshini Poola
asked Jan 31 at 18:56
Darshini PoolaDarshini Poola
126
asked Jan 31 at 18:56
Darshini PoolaDarshini Poola
126
asked Jan 31 at 18:56
Darshini PoolaDarshini Poola
126
126
$begingroup$
i get $$-frac{7}{4}pi,-frac{3}{4}pi,frac{pi}{4},frac{5}{4}pi$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:03
$begingroup$
So do I. I wanted to know whether I was making an error.
$endgroup$
– Darshini Poola
Jan 31 at 19:05
$begingroup$
You have no error in your calculations.
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:06
$begingroup$
So is the given solution incorrect?
$endgroup$
– Darshini Poola
Jan 31 at 19:07
$begingroup$
@Dr.SonnhardGraubner: If we assume to work in $mathbb R$, $log(cdot)$ is defined only when the argument is greater than $0$, which means that both sine and cosine must be postive, in this equation. A valid solution must be in the first quadrant. Correct?
$endgroup$
– Matteo
Jan 31 at 19:09
|
show 6 more comments
$begingroup$
i get $$-frac{7}{4}pi,-frac{3}{4}pi,frac{pi}{4},frac{5}{4}pi$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:03
$begingroup$
So do I. I wanted to know whether I was making an error.
$endgroup$
– Darshini Poola
Jan 31 at 19:05
$begingroup$
You have no error in your calculations.
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:06
$begingroup$
So is the given solution incorrect?
$endgroup$
– Darshini Poola
Jan 31 at 19:07
$begingroup$
@Dr.SonnhardGraubner: If we assume to work in $mathbb R$, $log(cdot)$ is defined only when the argument is greater than $0$, which means that both sine and cosine must be postive, in this equation. A valid solution must be in the first quadrant. Correct?
$endgroup$
– Matteo
Jan 31 at 19:09
$begingroup$
i get $$-frac{7}{4}pi,-frac{3}{4}pi,frac{pi}{4},frac{5}{4}pi$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:03
$begingroup$
i get $$-frac{7}{4}pi,-frac{3}{4}pi,frac{pi}{4},frac{5}{4}pi$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:03
$begingroup$
So do I. I wanted to know whether I was making an error.
$endgroup$
– Darshini Poola
Jan 31 at 19:05
$begingroup$
So do I. I wanted to know whether I was making an error.
$endgroup$
– Darshini Poola
Jan 31 at 19:05
$begingroup$
You have no error in your calculations.
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:06
$begingroup$
You have no error in your calculations.
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:06
$begingroup$
So is the given solution incorrect?
$endgroup$
– Darshini Poola
Jan 31 at 19:07
$begingroup$
So is the given solution incorrect?
$endgroup$
– Darshini Poola
Jan 31 at 19:07
$begingroup$
@Dr.SonnhardGraubner: If we assume to work in $mathbb R$, $log(cdot)$ is defined only when the argument is greater than $0$, which means that both sine and cosine must be postive, in this equation. A valid solution must be in the first quadrant. Correct?
$endgroup$
– Matteo
Jan 31 at 19:09
$begingroup$
@Dr.SonnhardGraubner: If we assume to work in $mathbb R$, $log(cdot)$ is defined only when the argument is greater than $0$, which means that both sine and cosine must be postive, in this equation. A valid solution must be in the first quadrant. Correct?
$endgroup$
– Matteo
Jan 31 at 19:09
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The book is giving one solution $[-2pi,2pi]$ but giving another $(0, pi/2)$. It looks like the book is inconsistent, but you are correct in your answer.
The general solution for the equation above is $$dfrac {pi}{4} + 2 pi k, k in (0, 1)$$.
EDIT: Removed the $pm$ from the previous solution as this would result in negative logarithms; expanded the domain.
answered Jan 31 at 19:23
bjcolby15bjcolby15
1,51711016
$endgroup$
$begingroup$
in the second quadrant $cos x$ is negative so the equation has no meaning.
$endgroup$
– Matteo
Jan 31 at 19:27
$begingroup$
Duly noted. I've corrected my post above and removed the $pm$ from the original solution.
$endgroup$
– bjcolby15
Jan 31 at 21:57
$begingroup$
the general solution is actually $frac{pi}{4} + 2kpi$, with $k in Bbb Z$, in order to always end up in the first quadrant. Then, if solutions in $[-2pi, 2pi]$ are required, those are $$-frac{7}{4}pi, frac{1}{4}pi$$.
$endgroup$
– Matteo
Jan 31 at 22:02
add a comment |
$begingroup$
Recall that in $Bbb R$ the property $log alpha + logbeta = logalphabeta$, is only valid when both $alpha$ and $beta$ are strictly positive.
The equation
$$log_{0.5}sin x=1-log_{0.5}cos x$$
is thus equivalent to
$$
begin{cases}
sin x> 0\
cos x > 0\
log_{0.5}sin xcos x = 1,
end{cases}
$$
meaning
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin x cos x = frac{1}{2}
end{cases} , kin Bbb Z
$$
or, equivalently,
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin 2x = 1.
end{cases}
$$
Solving with respect to $x$ the second equation yields
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
x = frac{pi}{4}+kpi
end{cases}
$$
The required solutions in $[-2pi, 2pi]$ therefore are $x = frac{pi}{4}$ and $x = frac{pi}{4}-2pi$.
answered Jan 31 at 19:13
MatteoMatteo
1,302313
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The book is giving one solution $[-2pi,2pi]$ but giving another $(0, pi/2)$. It looks like the book is inconsistent, but you are correct in your answer.
The general solution for the equation above is $$dfrac {pi}{4} + 2 pi k, k in (0, 1)$$.
EDIT: Removed the $pm$ from the previous solution as this would result in negative logarithms; expanded the domain.
answered Jan 31 at 19:23
bjcolby15bjcolby15
1,51711016
$endgroup$
$begingroup$
in the second quadrant $cos x$ is negative so the equation has no meaning.
$endgroup$
– Matteo
Jan 31 at 19:27
$begingroup$
Duly noted. I've corrected my post above and removed the $pm$ from the original solution.
$endgroup$
– bjcolby15
Jan 31 at 21:57
$begingroup$
the general solution is actually $frac{pi}{4} + 2kpi$, with $k in Bbb Z$, in order to always end up in the first quadrant. Then, if solutions in $[-2pi, 2pi]$ are required, those are $$-frac{7}{4}pi, frac{1}{4}pi$$.
$endgroup$
– Matteo
Jan 31 at 22:02
add a comment |
$begingroup$
The book is giving one solution $[-2pi,2pi]$ but giving another $(0, pi/2)$. It looks like the book is inconsistent, but you are correct in your answer.
The general solution for the equation above is $$dfrac {pi}{4} + 2 pi k, k in (0, 1)$$.
EDIT: Removed the $pm$ from the previous solution as this would result in negative logarithms; expanded the domain.
answered Jan 31 at 19:23
bjcolby15bjcolby15
1,51711016
$endgroup$
$begingroup$
in the second quadrant $cos x$ is negative so the equation has no meaning.
$endgroup$
– Matteo
Jan 31 at 19:27
$begingroup$
Duly noted. I've corrected my post above and removed the $pm$ from the original solution.
$endgroup$
– bjcolby15
Jan 31 at 21:57
$begingroup$
the general solution is actually $frac{pi}{4} + 2kpi$, with $k in Bbb Z$, in order to always end up in the first quadrant. Then, if solutions in $[-2pi, 2pi]$ are required, those are $$-frac{7}{4}pi, frac{1}{4}pi$$.
$endgroup$
– Matteo
Jan 31 at 22:02
add a comment |
$begingroup$
The book is giving one solution $[-2pi,2pi]$ but giving another $(0, pi/2)$. It looks like the book is inconsistent, but you are correct in your answer.
The general solution for the equation above is $$dfrac {pi}{4} + 2 pi k, k in (0, 1)$$.
EDIT: Removed the $pm$ from the previous solution as this would result in negative logarithms; expanded the domain.
answered Jan 31 at 19:23
bjcolby15bjcolby15
1,51711016
$endgroup$
The book is giving one solution $[-2pi,2pi]$ but giving another $(0, pi/2)$. It looks like the book is inconsistent, but you are correct in your answer.
The general solution for the equation above is $$dfrac {pi}{4} + 2 pi k, k in (0, 1)$$.
EDIT: Removed the $pm$ from the previous solution as this would result in negative logarithms; expanded the domain.
answered Jan 31 at 19:23
bjcolby15bjcolby15
1,51711016
edited Jan 31 at 22:19
answered Jan 31 at 19:23
bjcolby15bjcolby15
1,51711016
answered Jan 31 at 19:23
bjcolby15bjcolby15
1,51711016
answered Jan 31 at 19:23
bjcolby15bjcolby15
1,51711016
1,51711016
$begingroup$
in the second quadrant $cos x$ is negative so the equation has no meaning.
$endgroup$
– Matteo
Jan 31 at 19:27
$begingroup$
Duly noted. I've corrected my post above and removed the $pm$ from the original solution.
$endgroup$
– bjcolby15
Jan 31 at 21:57
$begingroup$
the general solution is actually $frac{pi}{4} + 2kpi$, with $k in Bbb Z$, in order to always end up in the first quadrant. Then, if solutions in $[-2pi, 2pi]$ are required, those are $$-frac{7}{4}pi, frac{1}{4}pi$$.
$endgroup$
– Matteo
Jan 31 at 22:02
add a comment |
$begingroup$
in the second quadrant $cos x$ is negative so the equation has no meaning.
$endgroup$
– Matteo
Jan 31 at 19:27
$begingroup$
Duly noted. I've corrected my post above and removed the $pm$ from the original solution.
$endgroup$
– bjcolby15
Jan 31 at 21:57
$begingroup$
the general solution is actually $frac{pi}{4} + 2kpi$, with $k in Bbb Z$, in order to always end up in the first quadrant. Then, if solutions in $[-2pi, 2pi]$ are required, those are $$-frac{7}{4}pi, frac{1}{4}pi$$.
$endgroup$
– Matteo
Jan 31 at 22:02
$begingroup$
in the second quadrant $cos x$ is negative so the equation has no meaning.
$endgroup$
– Matteo
Jan 31 at 19:27
$begingroup$
in the second quadrant $cos x$ is negative so the equation has no meaning.
$endgroup$
– Matteo
Jan 31 at 19:27
$begingroup$
Duly noted. I've corrected my post above and removed the $pm$ from the original solution.
$endgroup$
– bjcolby15
Jan 31 at 21:57
$begingroup$
Duly noted. I've corrected my post above and removed the $pm$ from the original solution.
$endgroup$
– bjcolby15
Jan 31 at 21:57
$begingroup$
the general solution is actually $frac{pi}{4} + 2kpi$, with $k in Bbb Z$, in order to always end up in the first quadrant. Then, if solutions in $[-2pi, 2pi]$ are required, those are $$-frac{7}{4}pi, frac{1}{4}pi$$.
$endgroup$
– Matteo
Jan 31 at 22:02
$begingroup$
the general solution is actually $frac{pi}{4} + 2kpi$, with $k in Bbb Z$, in order to always end up in the first quadrant. Then, if solutions in $[-2pi, 2pi]$ are required, those are $$-frac{7}{4}pi, frac{1}{4}pi$$.
$endgroup$
– Matteo
Jan 31 at 22:02
add a comment |
$begingroup$
Recall that in $Bbb R$ the property $log alpha + logbeta = logalphabeta$, is only valid when both $alpha$ and $beta$ are strictly positive.
The equation
$$log_{0.5}sin x=1-log_{0.5}cos x$$
is thus equivalent to
$$
begin{cases}
sin x> 0\
cos x > 0\
log_{0.5}sin xcos x = 1,
end{cases}
$$
meaning
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin x cos x = frac{1}{2}
end{cases} , kin Bbb Z
$$
or, equivalently,
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin 2x = 1.
end{cases}
$$
Solving with respect to $x$ the second equation yields
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
x = frac{pi}{4}+kpi
end{cases}
$$
The required solutions in $[-2pi, 2pi]$ therefore are $x = frac{pi}{4}$ and $x = frac{pi}{4}-2pi$.
answered Jan 31 at 19:13
MatteoMatteo
1,302313
$endgroup$
add a comment |
$begingroup$
Recall that in $Bbb R$ the property $log alpha + logbeta = logalphabeta$, is only valid when both $alpha$ and $beta$ are strictly positive.
The equation
$$log_{0.5}sin x=1-log_{0.5}cos x$$
is thus equivalent to
$$
begin{cases}
sin x> 0\
cos x > 0\
log_{0.5}sin xcos x = 1,
end{cases}
$$
meaning
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin x cos x = frac{1}{2}
end{cases} , kin Bbb Z
$$
or, equivalently,
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin 2x = 1.
end{cases}
$$
Solving with respect to $x$ the second equation yields
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
x = frac{pi}{4}+kpi
end{cases}
$$
The required solutions in $[-2pi, 2pi]$ therefore are $x = frac{pi}{4}$ and $x = frac{pi}{4}-2pi$.
answered Jan 31 at 19:13
MatteoMatteo
1,302313
$endgroup$
add a comment |
$begingroup$
Recall that in $Bbb R$ the property $log alpha + logbeta = logalphabeta$, is only valid when both $alpha$ and $beta$ are strictly positive.
The equation
$$log_{0.5}sin x=1-log_{0.5}cos x$$
is thus equivalent to
$$
begin{cases}
sin x> 0\
cos x > 0\
log_{0.5}sin xcos x = 1,
end{cases}
$$
meaning
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin x cos x = frac{1}{2}
end{cases} , kin Bbb Z
$$
or, equivalently,
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin 2x = 1.
end{cases}
$$
Solving with respect to $x$ the second equation yields
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
x = frac{pi}{4}+kpi
end{cases}
$$
The required solutions in $[-2pi, 2pi]$ therefore are $x = frac{pi}{4}$ and $x = frac{pi}{4}-2pi$.
answered Jan 31 at 19:13
MatteoMatteo
1,302313
$endgroup$
Recall that in $Bbb R$ the property $log alpha + logbeta = logalphabeta$, is only valid when both $alpha$ and $beta$ are strictly positive.
The equation
$$log_{0.5}sin x=1-log_{0.5}cos x$$
is thus equivalent to
$$
begin{cases}
sin x> 0\
cos x > 0\
log_{0.5}sin xcos x = 1,
end{cases}
$$
meaning
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin x cos x = frac{1}{2}
end{cases} , kin Bbb Z
$$
or, equivalently,
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
sin 2x = 1.
end{cases}
$$
Solving with respect to $x$ the second equation yields
$$
begin{cases}
2kpi < x < frac{pi}{2}+2kpi\
x = frac{pi}{4}+kpi
end{cases}
$$
The required solutions in $[-2pi, 2pi]$ therefore are $x = frac{pi}{4}$ and $x = frac{pi}{4}-2pi$.
answered Jan 31 at 19:13
MatteoMatteo
1,302313
edited Jan 31 at 22:08
answered Jan 31 at 19:13
MatteoMatteo
1,302313
answered Jan 31 at 19:13
MatteoMatteo
1,302313
answered Jan 31 at 19:13
MatteoMatteo
1,302313
1,302313
add a comment |
add a comment |
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$begingroup$
i get $$-frac{7}{4}pi,-frac{3}{4}pi,frac{pi}{4},frac{5}{4}pi$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:03
$begingroup$
So do I. I wanted to know whether I was making an error.
$endgroup$
– Darshini Poola
Jan 31 at 19:05
$begingroup$
You have no error in your calculations.
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 19:06
$begingroup$
So is the given solution incorrect?
$endgroup$
– Darshini Poola
Jan 31 at 19:07
$begingroup$
@Dr.SonnhardGraubner: If we assume to work in $mathbb R$, $log(cdot)$ is defined only when the argument is greater than $0$, which means that both sine and cosine must be postive, in this equation. A valid solution must be in the first quadrant. Correct?
$endgroup$
– Matteo
Jan 31 at 19:09