Losing solutions in $cos{x}+cos{2x}=0$
I'm trying to solve $cos x=-cos2x$, $-pi<xleqpi$, WITHOUT USING the double angle formulae, here's my current working:
$$begin{align*}cos x&=-cos2x\
cos (xpm2kpi)&=cos (2xpm(2k+1)pi)\
implies xpm2kpi&=2xpm(2k+1)pi\
x&=2xpmpi\
x&=pmpi\
therefore x&=pitext{ , in the given range of $x$}end{align*}$$
Graphically it's obvious I've missed two other solutions, $x=pmfrac{pi}{3}$, however I do not know why because I have not, to my knowledge, carried out any illegal operations such as dividing by zero.
Could someone point out my mistake please and give an edited solution? Thanks!
algebra-precalculus trigonometry proof-verification
asked Dec 12 '18 at 14:47
Adam BromileyAdam Bromiley
279110
add a comment |
I'm trying to solve $cos x=-cos2x$, $-pi<xleqpi$, WITHOUT USING the double angle formulae, here's my current working:
$$begin{align*}cos x&=-cos2x\
cos (xpm2kpi)&=cos (2xpm(2k+1)pi)\
implies xpm2kpi&=2xpm(2k+1)pi\
x&=2xpmpi\
x&=pmpi\
therefore x&=pitext{ , in the given range of $x$}end{align*}$$
Graphically it's obvious I've missed two other solutions, $x=pmfrac{pi}{3}$, however I do not know why because I have not, to my knowledge, carried out any illegal operations such as dividing by zero.
Could someone point out my mistake please and give an edited solution? Thanks!
algebra-precalculus trigonometry proof-verification
asked Dec 12 '18 at 14:47
Adam BromileyAdam Bromiley
279110
1
$k$ on the left and right sides should not be the same.
– Alex Silva
Dec 12 '18 at 14:50
@AlexSilva I graphed them on Desmos for integer values of $k$ and they always lined up with the original graphs? Or am I missing something?
– Adam Bromiley
Dec 12 '18 at 15:02
1
Your implication is wrong.
– Yves Daoust
Dec 12 '18 at 15:05
add a comment |
I'm trying to solve $cos x=-cos2x$, $-pi<xleqpi$, WITHOUT USING the double angle formulae, here's my current working:
$$begin{align*}cos x&=-cos2x\
cos (xpm2kpi)&=cos (2xpm(2k+1)pi)\
implies xpm2kpi&=2xpm(2k+1)pi\
x&=2xpmpi\
x&=pmpi\
therefore x&=pitext{ , in the given range of $x$}end{align*}$$
Graphically it's obvious I've missed two other solutions, $x=pmfrac{pi}{3}$, however I do not know why because I have not, to my knowledge, carried out any illegal operations such as dividing by zero.
Could someone point out my mistake please and give an edited solution? Thanks!
algebra-precalculus trigonometry proof-verification
asked Dec 12 '18 at 14:47
Adam BromileyAdam Bromiley
279110
I'm trying to solve $cos x=-cos2x$, $-pi<xleqpi$, WITHOUT USING the double angle formulae, here's my current working:
$$begin{align*}cos x&=-cos2x\
cos (xpm2kpi)&=cos (2xpm(2k+1)pi)\
implies xpm2kpi&=2xpm(2k+1)pi\
x&=2xpmpi\
x&=pmpi\
therefore x&=pitext{ , in the given range of $x$}end{align*}$$
Graphically it's obvious I've missed two other solutions, $x=pmfrac{pi}{3}$, however I do not know why because I have not, to my knowledge, carried out any illegal operations such as dividing by zero.
Could someone point out my mistake please and give an edited solution? Thanks!
algebra-precalculus trigonometry proof-verification
algebra-precalculus trigonometry proof-verification
asked Dec 12 '18 at 14:47
Adam BromileyAdam Bromiley
279110
asked Dec 12 '18 at 14:47
Adam BromileyAdam Bromiley
279110
asked Dec 12 '18 at 14:47
Adam BromileyAdam Bromiley
279110
asked Dec 12 '18 at 14:47
Adam BromileyAdam Bromiley
279110
asked Dec 12 '18 at 14:47
Adam BromileyAdam Bromiley
279110
279110
1
$k$ on the left and right sides should not be the same.
– Alex Silva
Dec 12 '18 at 14:50
@AlexSilva I graphed them on Desmos for integer values of $k$ and they always lined up with the original graphs? Or am I missing something?
– Adam Bromiley
Dec 12 '18 at 15:02
1
Your implication is wrong.
– Yves Daoust
Dec 12 '18 at 15:05
add a comment |
1
$k$ on the left and right sides should not be the same.
– Alex Silva
Dec 12 '18 at 14:50
@AlexSilva I graphed them on Desmos for integer values of $k$ and they always lined up with the original graphs? Or am I missing something?
– Adam Bromiley
Dec 12 '18 at 15:02
1
Your implication is wrong.
– Yves Daoust
Dec 12 '18 at 15:05
1
1
$k$ on the left and right sides should not be the same.
– Alex Silva
Dec 12 '18 at 14:50
$k$ on the left and right sides should not be the same.
– Alex Silva
Dec 12 '18 at 14:50
@AlexSilva I graphed them on Desmos for integer values of $k$ and they always lined up with the original graphs? Or am I missing something?
– Adam Bromiley
Dec 12 '18 at 15:02
@AlexSilva I graphed them on Desmos for integer values of $k$ and they always lined up with the original graphs? Or am I missing something?
– Adam Bromiley
Dec 12 '18 at 15:02
1
1
Your implication is wrong.
– Yves Daoust
Dec 12 '18 at 15:05
Your implication is wrong.
– Yves Daoust
Dec 12 '18 at 15:05
add a comment |
2 Answers
2
active
oldest
votes
Use that
$$cos x=-cos2x iff cos x=cos(pi-2x)$$
and that
$$cos alpha = cos theta iff alpha = theta +2kpi , lor , alpha = -theta +2kpi quad kin mathbb{Z}$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 12 '18 at 14:50
gimusigimusi
1
Ah okay, so from what I gathered (being still in highschool I'm not completely familiar with the if and only if implication so I've probably misinterpreted this answer), I came up with $pm x+2kpi = kpi pm 2x implies kpi = pm x,pm 3x$, which works but obviously only for odd-values of k, however k exists as an integer (and likewise works as an integer for my first statement in this comment)?
– Adam Bromiley
Dec 12 '18 at 15:17
@AdamBromiley The simbol $lor$ stays for logical "or".
– gimusi
Dec 12 '18 at 15:21
@AdamBromiley Then we have $$x=pi-2x+2kpi$$ or $$x=-(pi-2x)+2kpi$$ From here we can find all the solutions. Refer also to the given example. Do not hesitate to ask for any other clarification. Bye
– gimusi
Dec 12 '18 at 15:23
Ah okay I get that now, apart from the one implication on the first line of your answer, why is there not a $k$ in $cos(pi -2x)$? Surely it is true for any odd multiples of pi?
– Adam Bromiley
Dec 12 '18 at 15:26
@AdamBromiley The first line is always true, we use that to transform the given equation in an other equivalent in order to eliminate the minus sign in front. Then we can use the second identity to find the solutions.
– gimusi
Dec 12 '18 at 15:31
|
show 2 more comments
$$cos(x+pi)=cos(2x)iff x+pi=2kpipm2x.$$
Hence
$$x=(2k+1)pilor x=(2k+1)fracpi3.$$
answered Dec 12 '18 at 15:09
Yves DaoustYves Daoust
124k671222
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
Use that
$$cos x=-cos2x iff cos x=cos(pi-2x)$$
and that
$$cos alpha = cos theta iff alpha = theta +2kpi , lor , alpha = -theta +2kpi quad kin mathbb{Z}$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 12 '18 at 14:50
gimusigimusi
1
Ah okay, so from what I gathered (being still in highschool I'm not completely familiar with the if and only if implication so I've probably misinterpreted this answer), I came up with $pm x+2kpi = kpi pm 2x implies kpi = pm x,pm 3x$, which works but obviously only for odd-values of k, however k exists as an integer (and likewise works as an integer for my first statement in this comment)?
– Adam Bromiley
Dec 12 '18 at 15:17
@AdamBromiley The simbol $lor$ stays for logical "or".
– gimusi
Dec 12 '18 at 15:21
@AdamBromiley Then we have $$x=pi-2x+2kpi$$ or $$x=-(pi-2x)+2kpi$$ From here we can find all the solutions. Refer also to the given example. Do not hesitate to ask for any other clarification. Bye
– gimusi
Dec 12 '18 at 15:23
Ah okay I get that now, apart from the one implication on the first line of your answer, why is there not a $k$ in $cos(pi -2x)$? Surely it is true for any odd multiples of pi?
– Adam Bromiley
Dec 12 '18 at 15:26
@AdamBromiley The first line is always true, we use that to transform the given equation in an other equivalent in order to eliminate the minus sign in front. Then we can use the second identity to find the solutions.
– gimusi
Dec 12 '18 at 15:31
|
show 2 more comments
Use that
$$cos x=-cos2x iff cos x=cos(pi-2x)$$
and that
$$cos alpha = cos theta iff alpha = theta +2kpi , lor , alpha = -theta +2kpi quad kin mathbb{Z}$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 12 '18 at 14:50
gimusigimusi
1
Ah okay, so from what I gathered (being still in highschool I'm not completely familiar with the if and only if implication so I've probably misinterpreted this answer), I came up with $pm x+2kpi = kpi pm 2x implies kpi = pm x,pm 3x$, which works but obviously only for odd-values of k, however k exists as an integer (and likewise works as an integer for my first statement in this comment)?
– Adam Bromiley
Dec 12 '18 at 15:17
@AdamBromiley The simbol $lor$ stays for logical "or".
– gimusi
Dec 12 '18 at 15:21
@AdamBromiley Then we have $$x=pi-2x+2kpi$$ or $$x=-(pi-2x)+2kpi$$ From here we can find all the solutions. Refer also to the given example. Do not hesitate to ask for any other clarification. Bye
– gimusi
Dec 12 '18 at 15:23
Ah okay I get that now, apart from the one implication on the first line of your answer, why is there not a $k$ in $cos(pi -2x)$? Surely it is true for any odd multiples of pi?
– Adam Bromiley
Dec 12 '18 at 15:26
@AdamBromiley The first line is always true, we use that to transform the given equation in an other equivalent in order to eliminate the minus sign in front. Then we can use the second identity to find the solutions.
– gimusi
Dec 12 '18 at 15:31
|
show 2 more comments
Use that
$$cos x=-cos2x iff cos x=cos(pi-2x)$$
and that
$$cos alpha = cos theta iff alpha = theta +2kpi , lor , alpha = -theta +2kpi quad kin mathbb{Z}$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 12 '18 at 14:50
gimusigimusi
1
Use that
$$cos x=-cos2x iff cos x=cos(pi-2x)$$
and that
$$cos alpha = cos theta iff alpha = theta +2kpi , lor , alpha = -theta +2kpi quad kin mathbb{Z}$$
Refer also to the related
- Solving $cos(3x) = cos(2x)$
answered Dec 12 '18 at 14:50
gimusigimusi
1
edited Dec 12 '18 at 15:00
answered Dec 12 '18 at 14:50
gimusigimusi
1
answered Dec 12 '18 at 14:50
gimusigimusi
1
answered Dec 12 '18 at 14:50
gimusigimusi
1
1
Ah okay, so from what I gathered (being still in highschool I'm not completely familiar with the if and only if implication so I've probably misinterpreted this answer), I came up with $pm x+2kpi = kpi pm 2x implies kpi = pm x,pm 3x$, which works but obviously only for odd-values of k, however k exists as an integer (and likewise works as an integer for my first statement in this comment)?
– Adam Bromiley
Dec 12 '18 at 15:17
@AdamBromiley The simbol $lor$ stays for logical "or".
– gimusi
Dec 12 '18 at 15:21
@AdamBromiley Then we have $$x=pi-2x+2kpi$$ or $$x=-(pi-2x)+2kpi$$ From here we can find all the solutions. Refer also to the given example. Do not hesitate to ask for any other clarification. Bye
– gimusi
Dec 12 '18 at 15:23
Ah okay I get that now, apart from the one implication on the first line of your answer, why is there not a $k$ in $cos(pi -2x)$? Surely it is true for any odd multiples of pi?
– Adam Bromiley
Dec 12 '18 at 15:26
@AdamBromiley The first line is always true, we use that to transform the given equation in an other equivalent in order to eliminate the minus sign in front. Then we can use the second identity to find the solutions.
– gimusi
Dec 12 '18 at 15:31
|
show 2 more comments
Ah okay, so from what I gathered (being still in highschool I'm not completely familiar with the if and only if implication so I've probably misinterpreted this answer), I came up with $pm x+2kpi = kpi pm 2x implies kpi = pm x,pm 3x$, which works but obviously only for odd-values of k, however k exists as an integer (and likewise works as an integer for my first statement in this comment)?
– Adam Bromiley
Dec 12 '18 at 15:17
@AdamBromiley The simbol $lor$ stays for logical "or".
– gimusi
Dec 12 '18 at 15:21
@AdamBromiley Then we have $$x=pi-2x+2kpi$$ or $$x=-(pi-2x)+2kpi$$ From here we can find all the solutions. Refer also to the given example. Do not hesitate to ask for any other clarification. Bye
– gimusi
Dec 12 '18 at 15:23
Ah okay I get that now, apart from the one implication on the first line of your answer, why is there not a $k$ in $cos(pi -2x)$? Surely it is true for any odd multiples of pi?
– Adam Bromiley
Dec 12 '18 at 15:26
@AdamBromiley The first line is always true, we use that to transform the given equation in an other equivalent in order to eliminate the minus sign in front. Then we can use the second identity to find the solutions.
– gimusi
Dec 12 '18 at 15:31
Ah okay, so from what I gathered (being still in highschool I'm not completely familiar with the if and only if implication so I've probably misinterpreted this answer), I came up with $pm x+2kpi = kpi pm 2x implies kpi = pm x,pm 3x$, which works but obviously only for odd-values of k, however k exists as an integer (and likewise works as an integer for my first statement in this comment)?
– Adam Bromiley
Dec 12 '18 at 15:17
Ah okay, so from what I gathered (being still in highschool I'm not completely familiar with the if and only if implication so I've probably misinterpreted this answer), I came up with $pm x+2kpi = kpi pm 2x implies kpi = pm x,pm 3x$, which works but obviously only for odd-values of k, however k exists as an integer (and likewise works as an integer for my first statement in this comment)?
– Adam Bromiley
Dec 12 '18 at 15:17
@AdamBromiley The simbol $lor$ stays for logical "or".
– gimusi
Dec 12 '18 at 15:21
@AdamBromiley The simbol $lor$ stays for logical "or".
– gimusi
Dec 12 '18 at 15:21
@AdamBromiley Then we have $$x=pi-2x+2kpi$$ or $$x=-(pi-2x)+2kpi$$ From here we can find all the solutions. Refer also to the given example. Do not hesitate to ask for any other clarification. Bye
– gimusi
Dec 12 '18 at 15:23
@AdamBromiley Then we have $$x=pi-2x+2kpi$$ or $$x=-(pi-2x)+2kpi$$ From here we can find all the solutions. Refer also to the given example. Do not hesitate to ask for any other clarification. Bye
– gimusi
Dec 12 '18 at 15:23
Ah okay I get that now, apart from the one implication on the first line of your answer, why is there not a $k$ in $cos(pi -2x)$? Surely it is true for any odd multiples of pi?
– Adam Bromiley
Dec 12 '18 at 15:26
Ah okay I get that now, apart from the one implication on the first line of your answer, why is there not a $k$ in $cos(pi -2x)$? Surely it is true for any odd multiples of pi?
– Adam Bromiley
Dec 12 '18 at 15:26
@AdamBromiley The first line is always true, we use that to transform the given equation in an other equivalent in order to eliminate the minus sign in front. Then we can use the second identity to find the solutions.
– gimusi
Dec 12 '18 at 15:31
@AdamBromiley The first line is always true, we use that to transform the given equation in an other equivalent in order to eliminate the minus sign in front. Then we can use the second identity to find the solutions.
– gimusi
Dec 12 '18 at 15:31
|
show 2 more comments
$$cos(x+pi)=cos(2x)iff x+pi=2kpipm2x.$$
Hence
$$x=(2k+1)pilor x=(2k+1)fracpi3.$$
answered Dec 12 '18 at 15:09
Yves DaoustYves Daoust
124k671222
add a comment |
$$cos(x+pi)=cos(2x)iff x+pi=2kpipm2x.$$
Hence
$$x=(2k+1)pilor x=(2k+1)fracpi3.$$
answered Dec 12 '18 at 15:09
Yves DaoustYves Daoust
124k671222
add a comment |
$$cos(x+pi)=cos(2x)iff x+pi=2kpipm2x.$$
Hence
$$x=(2k+1)pilor x=(2k+1)fracpi3.$$
answered Dec 12 '18 at 15:09
Yves DaoustYves Daoust
124k671222
$$cos(x+pi)=cos(2x)iff x+pi=2kpipm2x.$$
Hence
$$x=(2k+1)pilor x=(2k+1)fracpi3.$$
answered Dec 12 '18 at 15:09
Yves DaoustYves Daoust
124k671222
answered Dec 12 '18 at 15:09
Yves DaoustYves Daoust
124k671222
answered Dec 12 '18 at 15:09
Yves DaoustYves Daoust
124k671222
answered Dec 12 '18 at 15:09
Yves DaoustYves Daoust
124k671222
124k671222
add a comment |
add a comment |
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1
$k$ on the left and right sides should not be the same.
– Alex Silva
Dec 12 '18 at 14:50
@AlexSilva I graphed them on Desmos for integer values of $k$ and they always lined up with the original graphs? Or am I missing something?
– Adam Bromiley
Dec 12 '18 at 15:02
1
Your implication is wrong.
– Yves Daoust
Dec 12 '18 at 15:05