Solve $cos^n x -sin^n x =1$ for $x$
$begingroup$
I already post a question on the solution of $$cos^n x +sin^n x =1,$$ but it's just a mistake. My real question is $$cos^n x -sin^n x =1.$$
calculus real-analysis
edited Dec 26 '18 at 2:30
Rócherz
2,7762721
asked Jul 29 '16 at 0:22
infox09infox09
105
$endgroup$
|
show 1 more comment
$begingroup$
I already post a question on the solution of $$cos^n x +sin^n x =1,$$ but it's just a mistake. My real question is $$cos^n x -sin^n x =1.$$
calculus real-analysis
edited Dec 26 '18 at 2:30
Rócherz
2,7762721
asked Jul 29 '16 at 0:22
infox09infox09
105
$endgroup$
$begingroup$
Please don't use the align environment in titles.
$endgroup$
– MCT
Jul 29 '16 at 0:55
$begingroup$
it changes nothing, see our answers
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
Are you interested in solving for $n$ or for $x$? Is $n$ an integer?
$endgroup$
– Joel Reyes Noche
Jul 29 '16 at 1:02
$begingroup$
surr for x , n just is integer parameter
$endgroup$
– infox09
Jul 29 '16 at 1:05
$begingroup$
The problem is easy for $n$ a positive integer. For negative integers the situation is different.
$endgroup$
– André Nicolas
Jul 29 '16 at 1:43
|
show 1 more comment
$begingroup$
I already post a question on the solution of $$cos^n x +sin^n x =1,$$ but it's just a mistake. My real question is $$cos^n x -sin^n x =1.$$
calculus real-analysis
edited Dec 26 '18 at 2:30
Rócherz
2,7762721
asked Jul 29 '16 at 0:22
infox09infox09
105
$endgroup$
I already post a question on the solution of $$cos^n x +sin^n x =1,$$ but it's just a mistake. My real question is $$cos^n x -sin^n x =1.$$
calculus real-analysis
calculus real-analysis
edited Dec 26 '18 at 2:30
Rócherz
2,7762721
asked Jul 29 '16 at 0:22
infox09infox09
105
edited Dec 26 '18 at 2:30
Rócherz
2,7762721
asked Jul 29 '16 at 0:22
infox09infox09
105
edited Dec 26 '18 at 2:30
Rócherz
2,7762721
edited Dec 26 '18 at 2:30
Rócherz
2,7762721
edited Dec 26 '18 at 2:30
Rócherz
2,7762721
2,7762721
asked Jul 29 '16 at 0:22
infox09infox09
105
asked Jul 29 '16 at 0:22
infox09infox09
105
asked Jul 29 '16 at 0:22
infox09infox09
105
105
$begingroup$
Please don't use the align environment in titles.
$endgroup$
– MCT
Jul 29 '16 at 0:55
$begingroup$
it changes nothing, see our answers
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
Are you interested in solving for $n$ or for $x$? Is $n$ an integer?
$endgroup$
– Joel Reyes Noche
Jul 29 '16 at 1:02
$begingroup$
surr for x , n just is integer parameter
$endgroup$
– infox09
Jul 29 '16 at 1:05
$begingroup$
The problem is easy for $n$ a positive integer. For negative integers the situation is different.
$endgroup$
– André Nicolas
Jul 29 '16 at 1:43
|
show 1 more comment
$begingroup$
Please don't use the align environment in titles.
$endgroup$
– MCT
Jul 29 '16 at 0:55
$begingroup$
it changes nothing, see our answers
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
Are you interested in solving for $n$ or for $x$? Is $n$ an integer?
$endgroup$
– Joel Reyes Noche
Jul 29 '16 at 1:02
$begingroup$
surr for x , n just is integer parameter
$endgroup$
– infox09
Jul 29 '16 at 1:05
$begingroup$
The problem is easy for $n$ a positive integer. For negative integers the situation is different.
$endgroup$
– André Nicolas
Jul 29 '16 at 1:43
$begingroup$
Please don't use the align environment in titles.
$endgroup$
– MCT
Jul 29 '16 at 0:55
$begingroup$
Please don't use the align environment in titles.
$endgroup$
– MCT
Jul 29 '16 at 0:55
$begingroup$
it changes nothing, see our answers
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
it changes nothing, see our answers
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
Are you interested in solving for $n$ or for $x$? Is $n$ an integer?
$endgroup$
– Joel Reyes Noche
Jul 29 '16 at 1:02
$begingroup$
Are you interested in solving for $n$ or for $x$? Is $n$ an integer?
$endgroup$
– Joel Reyes Noche
Jul 29 '16 at 1:02
$begingroup$
surr for x , n just is integer parameter
$endgroup$
– infox09
Jul 29 '16 at 1:05
$begingroup$
surr for x , n just is integer parameter
$endgroup$
– infox09
Jul 29 '16 at 1:05
$begingroup$
The problem is easy for $n$ a positive integer. For negative integers the situation is different.
$endgroup$
– André Nicolas
Jul 29 '16 at 1:43
$begingroup$
The problem is easy for $n$ a positive integer. For negative integers the situation is different.
$endgroup$
– André Nicolas
Jul 29 '16 at 1:43
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If $n$ is odd, your equation is just:
$$cos^n (-x) + sin^n (-x) = 1$$
reducing the problem to one already solved.
Suppose that $n$ is even. We know that $cos^n(x) le 1$. If $sin (x) neq 0$, then $sin^n(x) > 0$, hence $cos^n (x) - sin^n(x) < 1$, so there are no solutions. If $sin (x) = 0$, then $x in pi Bbb Z$, so $cos^2(x) = 1$ and the equation holds.
Summary: for even $n$, the set of solutions is ${pi k, k in Bbb Z}$.
$endgroup$
$begingroup$
yes if x is odd, the solution is clear but the other case please a small detail
$endgroup$
– infox09
Jul 29 '16 at 0:34
$begingroup$
@infox09 no the solution isn't clear for you
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
thank u ahmed now it is clair
$endgroup$
– infox09
Jul 29 '16 at 1:11
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $n$ is odd, your equation is just:
$$cos^n (-x) + sin^n (-x) = 1$$
reducing the problem to one already solved.
Suppose that $n$ is even. We know that $cos^n(x) le 1$. If $sin (x) neq 0$, then $sin^n(x) > 0$, hence $cos^n (x) - sin^n(x) < 1$, so there are no solutions. If $sin (x) = 0$, then $x in pi Bbb Z$, so $cos^2(x) = 1$ and the equation holds.
Summary: for even $n$, the set of solutions is ${pi k, k in Bbb Z}$.
$endgroup$
$begingroup$
yes if x is odd, the solution is clear but the other case please a small detail
$endgroup$
– infox09
Jul 29 '16 at 0:34
$begingroup$
@infox09 no the solution isn't clear for you
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
thank u ahmed now it is clair
$endgroup$
– infox09
Jul 29 '16 at 1:11
add a comment |
$begingroup$
If $n$ is odd, your equation is just:
$$cos^n (-x) + sin^n (-x) = 1$$
reducing the problem to one already solved.
Suppose that $n$ is even. We know that $cos^n(x) le 1$. If $sin (x) neq 0$, then $sin^n(x) > 0$, hence $cos^n (x) - sin^n(x) < 1$, so there are no solutions. If $sin (x) = 0$, then $x in pi Bbb Z$, so $cos^2(x) = 1$ and the equation holds.
Summary: for even $n$, the set of solutions is ${pi k, k in Bbb Z}$.
$endgroup$
$begingroup$
yes if x is odd, the solution is clear but the other case please a small detail
$endgroup$
– infox09
Jul 29 '16 at 0:34
$begingroup$
@infox09 no the solution isn't clear for you
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
thank u ahmed now it is clair
$endgroup$
– infox09
Jul 29 '16 at 1:11
add a comment |
$begingroup$
If $n$ is odd, your equation is just:
$$cos^n (-x) + sin^n (-x) = 1$$
reducing the problem to one already solved.
Suppose that $n$ is even. We know that $cos^n(x) le 1$. If $sin (x) neq 0$, then $sin^n(x) > 0$, hence $cos^n (x) - sin^n(x) < 1$, so there are no solutions. If $sin (x) = 0$, then $x in pi Bbb Z$, so $cos^2(x) = 1$ and the equation holds.
Summary: for even $n$, the set of solutions is ${pi k, k in Bbb Z}$.
$endgroup$
If $n$ is odd, your equation is just:
$$cos^n (-x) + sin^n (-x) = 1$$
reducing the problem to one already solved.
Suppose that $n$ is even. We know that $cos^n(x) le 1$. If $sin (x) neq 0$, then $sin^n(x) > 0$, hence $cos^n (x) - sin^n(x) < 1$, so there are no solutions. If $sin (x) = 0$, then $x in pi Bbb Z$, so $cos^2(x) = 1$ and the equation holds.
Summary: for even $n$, the set of solutions is ${pi k, k in Bbb Z}$.
edited Jul 29 '16 at 0:38
answered Jul 29 '16 at 0:27
user258700
$begingroup$
yes if x is odd, the solution is clear but the other case please a small detail
$endgroup$
– infox09
Jul 29 '16 at 0:34
$begingroup$
@infox09 no the solution isn't clear for you
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
thank u ahmed now it is clair
$endgroup$
– infox09
Jul 29 '16 at 1:11
add a comment |
$begingroup$
yes if x is odd, the solution is clear but the other case please a small detail
$endgroup$
– infox09
Jul 29 '16 at 0:34
$begingroup$
@infox09 no the solution isn't clear for you
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
thank u ahmed now it is clair
$endgroup$
– infox09
Jul 29 '16 at 1:11
$begingroup$
yes if x is odd, the solution is clear but the other case please a small detail
$endgroup$
– infox09
Jul 29 '16 at 0:34
$begingroup$
yes if x is odd, the solution is clear but the other case please a small detail
$endgroup$
– infox09
Jul 29 '16 at 0:34
$begingroup$
@infox09 no the solution isn't clear for you
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
@infox09 no the solution isn't clear for you
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
thank u ahmed now it is clair
$endgroup$
– infox09
Jul 29 '16 at 1:11
$begingroup$
thank u ahmed now it is clair
$endgroup$
– infox09
Jul 29 '16 at 1:11
add a comment |
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$begingroup$
Please don't use the align environment in titles.
$endgroup$
– MCT
Jul 29 '16 at 0:55
$begingroup$
it changes nothing, see our answers
$endgroup$
– reuns
Jul 29 '16 at 1:00
$begingroup$
Are you interested in solving for $n$ or for $x$? Is $n$ an integer?
$endgroup$
– Joel Reyes Noche
Jul 29 '16 at 1:02
$begingroup$
surr for x , n just is integer parameter
$endgroup$
– infox09
Jul 29 '16 at 1:05
$begingroup$
The problem is easy for $n$ a positive integer. For negative integers the situation is different.
$endgroup$
– André Nicolas
Jul 29 '16 at 1:43