Solve $sqrt{2}sec x+tan x=1$
$begingroup$
Solve $sqrt{2}sec x+tan x=1$
I understand it can be very easily solved by expanding in terms of $sin x$ and $cos x$, gives $x=2npi-frac{pi}{4}$. But, what if I do the following:
$$
sqrt{2}sec x+tan x=1\
text{Differentiating}impliessqrt{2}sec xtan x+sec^2 x=0impliessqrt{2}tan x+sec x=0
$$
Step 1
$$
sec x=-sqrt{2}tan x=frac{1-tan x}{sqrt{2}}implies2tan x=tan x-1impliestan x=-1\
boxed{x=npi-frac{pi}{4}}
$$
Step 2
$$
tan x=1-sqrt{2}sec x=frac{-sec x}{sqrt{2}}implies2sec x-sqrt{2}=sec x\
impliessec x=sqrt{2}impliesboxed{x=2npipmfrac{pi}{4}}
$$
$$
x=npi-frac{pi}{4}quad&quad x=2npipmfrac{pi}{4}\implies bigg[x=2npi-frac{pi}{4}text{ or }x=2npi+frac{3pi}{4}bigg]quad&quad x=2npipmfrac{pi}{4}\
implies boxed{x=2npi-frac{pi}{4}}
$$
In my attempt why do I need Step 2 to get the complete solution ?
Can someone give a proper explanation to my attempt ?
trigonometry
asked Dec 3 '18 at 10:01
ss1729ss1729
1,9671923
$endgroup$
add a comment |
$begingroup$
Solve $sqrt{2}sec x+tan x=1$
I understand it can be very easily solved by expanding in terms of $sin x$ and $cos x$, gives $x=2npi-frac{pi}{4}$. But, what if I do the following:
$$
sqrt{2}sec x+tan x=1\
text{Differentiating}impliessqrt{2}sec xtan x+sec^2 x=0impliessqrt{2}tan x+sec x=0
$$
Step 1
$$
sec x=-sqrt{2}tan x=frac{1-tan x}{sqrt{2}}implies2tan x=tan x-1impliestan x=-1\
boxed{x=npi-frac{pi}{4}}
$$
Step 2
$$
tan x=1-sqrt{2}sec x=frac{-sec x}{sqrt{2}}implies2sec x-sqrt{2}=sec x\
impliessec x=sqrt{2}impliesboxed{x=2npipmfrac{pi}{4}}
$$
$$
x=npi-frac{pi}{4}quad&quad x=2npipmfrac{pi}{4}\implies bigg[x=2npi-frac{pi}{4}text{ or }x=2npi+frac{3pi}{4}bigg]quad&quad x=2npipmfrac{pi}{4}\
implies boxed{x=2npi-frac{pi}{4}}
$$
In my attempt why do I need Step 2 to get the complete solution ?
Can someone give a proper explanation to my attempt ?
trigonometry
asked Dec 3 '18 at 10:01
ss1729ss1729
1,9671923
$endgroup$
6
$begingroup$
This is not an answer to your question, but I'd note that the differentiation step is not valid. Try to apply it to something like $x^2 + 1 = 2$. Differentiating should give $2x = 0$, so $x = 0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are. I am not sure why it seems to work out in this case.
$endgroup$
– platty
Dec 3 '18 at 10:12
$begingroup$
Please let me know if my answer with an alternative method could be useful, otherwise I'll delete that.
$endgroup$
– gimusi
Dec 3 '18 at 10:49
2
$begingroup$
Ah, I think I see why it happens to work out. It turns out that at $2 pi n - pi/4$, the equation $sqrt{2} tan x + sec x = 0$ is actually true. So in a way, you end up using the fact that your answer works to derive it, in a sort of circular argument. Of course, like I mentioned above, this method is invalid, but now we know why it appears to work.
$endgroup$
– platty
Dec 3 '18 at 10:59
2
$begingroup$
@platty thanx very much. you could post it as answer though !
$endgroup$
– ss1729
Dec 3 '18 at 13:33
add a comment |
$begingroup$
Solve $sqrt{2}sec x+tan x=1$
I understand it can be very easily solved by expanding in terms of $sin x$ and $cos x$, gives $x=2npi-frac{pi}{4}$. But, what if I do the following:
$$
sqrt{2}sec x+tan x=1\
text{Differentiating}impliessqrt{2}sec xtan x+sec^2 x=0impliessqrt{2}tan x+sec x=0
$$
Step 1
$$
sec x=-sqrt{2}tan x=frac{1-tan x}{sqrt{2}}implies2tan x=tan x-1impliestan x=-1\
boxed{x=npi-frac{pi}{4}}
$$
Step 2
$$
tan x=1-sqrt{2}sec x=frac{-sec x}{sqrt{2}}implies2sec x-sqrt{2}=sec x\
impliessec x=sqrt{2}impliesboxed{x=2npipmfrac{pi}{4}}
$$
$$
x=npi-frac{pi}{4}quad&quad x=2npipmfrac{pi}{4}\implies bigg[x=2npi-frac{pi}{4}text{ or }x=2npi+frac{3pi}{4}bigg]quad&quad x=2npipmfrac{pi}{4}\
implies boxed{x=2npi-frac{pi}{4}}
$$
In my attempt why do I need Step 2 to get the complete solution ?
Can someone give a proper explanation to my attempt ?
trigonometry
asked Dec 3 '18 at 10:01
ss1729ss1729
1,9671923
$endgroup$
Solve $sqrt{2}sec x+tan x=1$
I understand it can be very easily solved by expanding in terms of $sin x$ and $cos x$, gives $x=2npi-frac{pi}{4}$. But, what if I do the following:
$$
sqrt{2}sec x+tan x=1\
text{Differentiating}impliessqrt{2}sec xtan x+sec^2 x=0impliessqrt{2}tan x+sec x=0
$$
Step 1
$$
sec x=-sqrt{2}tan x=frac{1-tan x}{sqrt{2}}implies2tan x=tan x-1impliestan x=-1\
boxed{x=npi-frac{pi}{4}}
$$
Step 2
$$
tan x=1-sqrt{2}sec x=frac{-sec x}{sqrt{2}}implies2sec x-sqrt{2}=sec x\
impliessec x=sqrt{2}impliesboxed{x=2npipmfrac{pi}{4}}
$$
$$
x=npi-frac{pi}{4}quad&quad x=2npipmfrac{pi}{4}\implies bigg[x=2npi-frac{pi}{4}text{ or }x=2npi+frac{3pi}{4}bigg]quad&quad x=2npipmfrac{pi}{4}\
implies boxed{x=2npi-frac{pi}{4}}
$$
In my attempt why do I need Step 2 to get the complete solution ?
Can someone give a proper explanation to my attempt ?
trigonometry
trigonometry
asked Dec 3 '18 at 10:01
ss1729ss1729
1,9671923
asked Dec 3 '18 at 10:01
ss1729ss1729
1,9671923
edited Dec 3 '18 at 10:19
ss1729
asked Dec 3 '18 at 10:01
ss1729ss1729
1,9671923
asked Dec 3 '18 at 10:01
ss1729ss1729
1,9671923
asked Dec 3 '18 at 10:01
ss1729ss1729
1,9671923
1,9671923
6
$begingroup$
This is not an answer to your question, but I'd note that the differentiation step is not valid. Try to apply it to something like $x^2 + 1 = 2$. Differentiating should give $2x = 0$, so $x = 0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are. I am not sure why it seems to work out in this case.
$endgroup$
– platty
Dec 3 '18 at 10:12
$begingroup$
Please let me know if my answer with an alternative method could be useful, otherwise I'll delete that.
$endgroup$
– gimusi
Dec 3 '18 at 10:49
2
$begingroup$
Ah, I think I see why it happens to work out. It turns out that at $2 pi n - pi/4$, the equation $sqrt{2} tan x + sec x = 0$ is actually true. So in a way, you end up using the fact that your answer works to derive it, in a sort of circular argument. Of course, like I mentioned above, this method is invalid, but now we know why it appears to work.
$endgroup$
– platty
Dec 3 '18 at 10:59
2
$begingroup$
@platty thanx very much. you could post it as answer though !
$endgroup$
– ss1729
Dec 3 '18 at 13:33
add a comment |
6
$begingroup$
This is not an answer to your question, but I'd note that the differentiation step is not valid. Try to apply it to something like $x^2 + 1 = 2$. Differentiating should give $2x = 0$, so $x = 0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are. I am not sure why it seems to work out in this case.
$endgroup$
– platty
Dec 3 '18 at 10:12
$begingroup$
Please let me know if my answer with an alternative method could be useful, otherwise I'll delete that.
$endgroup$
– gimusi
Dec 3 '18 at 10:49
2
$begingroup$
Ah, I think I see why it happens to work out. It turns out that at $2 pi n - pi/4$, the equation $sqrt{2} tan x + sec x = 0$ is actually true. So in a way, you end up using the fact that your answer works to derive it, in a sort of circular argument. Of course, like I mentioned above, this method is invalid, but now we know why it appears to work.
$endgroup$
– platty
Dec 3 '18 at 10:59
2
$begingroup$
@platty thanx very much. you could post it as answer though !
$endgroup$
– ss1729
Dec 3 '18 at 13:33
6
6
$begingroup$
This is not an answer to your question, but I'd note that the differentiation step is not valid. Try to apply it to something like $x^2 + 1 = 2$. Differentiating should give $2x = 0$, so $x = 0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are. I am not sure why it seems to work out in this case.
$endgroup$
– platty
Dec 3 '18 at 10:12
$begingroup$
This is not an answer to your question, but I'd note that the differentiation step is not valid. Try to apply it to something like $x^2 + 1 = 2$. Differentiating should give $2x = 0$, so $x = 0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are. I am not sure why it seems to work out in this case.
$endgroup$
– platty
Dec 3 '18 at 10:12
$begingroup$
Please let me know if my answer with an alternative method could be useful, otherwise I'll delete that.
$endgroup$
– gimusi
Dec 3 '18 at 10:49
$begingroup$
Please let me know if my answer with an alternative method could be useful, otherwise I'll delete that.
$endgroup$
– gimusi
Dec 3 '18 at 10:49
2
2
$begingroup$
Ah, I think I see why it happens to work out. It turns out that at $2 pi n - pi/4$, the equation $sqrt{2} tan x + sec x = 0$ is actually true. So in a way, you end up using the fact that your answer works to derive it, in a sort of circular argument. Of course, like I mentioned above, this method is invalid, but now we know why it appears to work.
$endgroup$
– platty
Dec 3 '18 at 10:59
$begingroup$
Ah, I think I see why it happens to work out. It turns out that at $2 pi n - pi/4$, the equation $sqrt{2} tan x + sec x = 0$ is actually true. So in a way, you end up using the fact that your answer works to derive it, in a sort of circular argument. Of course, like I mentioned above, this method is invalid, but now we know why it appears to work.
$endgroup$
– platty
Dec 3 '18 at 10:59
2
2
$begingroup$
@platty thanx very much. you could post it as answer though !
$endgroup$
– ss1729
Dec 3 '18 at 13:33
$begingroup$
@platty thanx very much. you could post it as answer though !
$endgroup$
– ss1729
Dec 3 '18 at 13:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The differentiation step is invalid. Try to apply it to something like $x^2+1=2$. Differentiating should give $2x=0$, so $x=0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are.
It turns out that, by coincidence, $sqrt{2} tanleft(2npi - frac{pi}{4} right) + secleft(2npi - frac{pi}{4} right) = 0$, i.e. $sqrt{2} tan x + sec x = 0$ does hold at the particular value of $x$ you want to find. As such, your manipulations are technically "valid" in that everything you say is true, but only because the answer that you are trying to find makes these expressions true. The argument, however, is circular, so the proof fails (you are starting by presuming what you want to show, i.e. $x = 2 pi n - frac{pi}{4}$). If, for example, the original question were $sqrt{2} sec x + tan x = 0$, it should be apparent that this method fails.
answered Dec 3 '18 at 19:13
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
HINT
As noticed we can't use differentiation to obtain the result indeed in general
$$f'(x)=g'(x) not Rightarrow f(x)=g(x)$$
consider for example the simple case
$$2x+3=1 to 2=0$$
I suggest to use tangent half-angle identities by $t = tan frac x2$ to obtain
$$sqrt{2}sec x+tan x=1 iff sqrt{2}frac{1+t^2}{1-t^2}+frac{2t}{1-t^2}=1$$
answered Dec 3 '18 at 10:20
gimusigimusi
92.8k84494
$endgroup$
3
$begingroup$
Is this an answer to the question “Can someone give a proper explanation to my attempt ?” ?
$endgroup$
– Martin R
Dec 3 '18 at 10:39
$begingroup$
@MartinR Yes you are right, I add some comment on that.
$endgroup$
– gimusi
Dec 3 '18 at 10:40
$begingroup$
No, but it’s a hint as labelled by ‘HINT’
$endgroup$
– DavidG
Dec 3 '18 at 10:41
2
$begingroup$
@DavidG: OP explicitly states that (s)he knows how to solve the equation, but has a problem with an alternative approach, and that is what the question is about. I would expect that a >80K user reads the whole question before posting, and answered accordingly
$endgroup$
– Martin R
Dec 3 '18 at 10:43
$begingroup$
Fair enough. Mistakes do happen.
$endgroup$
– DavidG
Dec 3 '18 at 10:47
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023863%2fsolve-sqrt2-sec-x-tan-x-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The differentiation step is invalid. Try to apply it to something like $x^2+1=2$. Differentiating should give $2x=0$, so $x=0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are.
It turns out that, by coincidence, $sqrt{2} tanleft(2npi - frac{pi}{4} right) + secleft(2npi - frac{pi}{4} right) = 0$, i.e. $sqrt{2} tan x + sec x = 0$ does hold at the particular value of $x$ you want to find. As such, your manipulations are technically "valid" in that everything you say is true, but only because the answer that you are trying to find makes these expressions true. The argument, however, is circular, so the proof fails (you are starting by presuming what you want to show, i.e. $x = 2 pi n - frac{pi}{4}$). If, for example, the original question were $sqrt{2} sec x + tan x = 0$, it should be apparent that this method fails.
answered Dec 3 '18 at 19:13
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
The differentiation step is invalid. Try to apply it to something like $x^2+1=2$. Differentiating should give $2x=0$, so $x=0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are.
It turns out that, by coincidence, $sqrt{2} tanleft(2npi - frac{pi}{4} right) + secleft(2npi - frac{pi}{4} right) = 0$, i.e. $sqrt{2} tan x + sec x = 0$ does hold at the particular value of $x$ you want to find. As such, your manipulations are technically "valid" in that everything you say is true, but only because the answer that you are trying to find makes these expressions true. The argument, however, is circular, so the proof fails (you are starting by presuming what you want to show, i.e. $x = 2 pi n - frac{pi}{4}$). If, for example, the original question were $sqrt{2} sec x + tan x = 0$, it should be apparent that this method fails.
answered Dec 3 '18 at 19:13
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
The differentiation step is invalid. Try to apply it to something like $x^2+1=2$. Differentiating should give $2x=0$, so $x=0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are.
It turns out that, by coincidence, $sqrt{2} tanleft(2npi - frac{pi}{4} right) + secleft(2npi - frac{pi}{4} right) = 0$, i.e. $sqrt{2} tan x + sec x = 0$ does hold at the particular value of $x$ you want to find. As such, your manipulations are technically "valid" in that everything you say is true, but only because the answer that you are trying to find makes these expressions true. The argument, however, is circular, so the proof fails (you are starting by presuming what you want to show, i.e. $x = 2 pi n - frac{pi}{4}$). If, for example, the original question were $sqrt{2} sec x + tan x = 0$, it should be apparent that this method fails.
answered Dec 3 '18 at 19:13
plattyplatty
3,370320
$endgroup$
The differentiation step is invalid. Try to apply it to something like $x^2+1=2$. Differentiating should give $2x=0$, so $x=0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are.
It turns out that, by coincidence, $sqrt{2} tanleft(2npi - frac{pi}{4} right) + secleft(2npi - frac{pi}{4} right) = 0$, i.e. $sqrt{2} tan x + sec x = 0$ does hold at the particular value of $x$ you want to find. As such, your manipulations are technically "valid" in that everything you say is true, but only because the answer that you are trying to find makes these expressions true. The argument, however, is circular, so the proof fails (you are starting by presuming what you want to show, i.e. $x = 2 pi n - frac{pi}{4}$). If, for example, the original question were $sqrt{2} sec x + tan x = 0$, it should be apparent that this method fails.
answered Dec 3 '18 at 19:13
plattyplatty
3,370320
answered Dec 3 '18 at 19:13
plattyplatty
3,370320
answered Dec 3 '18 at 19:13
plattyplatty
3,370320
answered Dec 3 '18 at 19:13
plattyplatty
3,370320
3,370320
add a comment |
add a comment |
$begingroup$
HINT
As noticed we can't use differentiation to obtain the result indeed in general
$$f'(x)=g'(x) not Rightarrow f(x)=g(x)$$
consider for example the simple case
$$2x+3=1 to 2=0$$
I suggest to use tangent half-angle identities by $t = tan frac x2$ to obtain
$$sqrt{2}sec x+tan x=1 iff sqrt{2}frac{1+t^2}{1-t^2}+frac{2t}{1-t^2}=1$$
answered Dec 3 '18 at 10:20
gimusigimusi
92.8k84494
$endgroup$
3
$begingroup$
Is this an answer to the question “Can someone give a proper explanation to my attempt ?” ?
$endgroup$
– Martin R
Dec 3 '18 at 10:39
$begingroup$
@MartinR Yes you are right, I add some comment on that.
$endgroup$
– gimusi
Dec 3 '18 at 10:40
$begingroup$
No, but it’s a hint as labelled by ‘HINT’
$endgroup$
– DavidG
Dec 3 '18 at 10:41
2
$begingroup$
@DavidG: OP explicitly states that (s)he knows how to solve the equation, but has a problem with an alternative approach, and that is what the question is about. I would expect that a >80K user reads the whole question before posting, and answered accordingly
$endgroup$
– Martin R
Dec 3 '18 at 10:43
$begingroup$
Fair enough. Mistakes do happen.
$endgroup$
– DavidG
Dec 3 '18 at 10:47
|
show 1 more comment
$begingroup$
HINT
As noticed we can't use differentiation to obtain the result indeed in general
$$f'(x)=g'(x) not Rightarrow f(x)=g(x)$$
consider for example the simple case
$$2x+3=1 to 2=0$$
I suggest to use tangent half-angle identities by $t = tan frac x2$ to obtain
$$sqrt{2}sec x+tan x=1 iff sqrt{2}frac{1+t^2}{1-t^2}+frac{2t}{1-t^2}=1$$
answered Dec 3 '18 at 10:20
gimusigimusi
92.8k84494
$endgroup$
3
$begingroup$
Is this an answer to the question “Can someone give a proper explanation to my attempt ?” ?
$endgroup$
– Martin R
Dec 3 '18 at 10:39
$begingroup$
@MartinR Yes you are right, I add some comment on that.
$endgroup$
– gimusi
Dec 3 '18 at 10:40
$begingroup$
No, but it’s a hint as labelled by ‘HINT’
$endgroup$
– DavidG
Dec 3 '18 at 10:41
2
$begingroup$
@DavidG: OP explicitly states that (s)he knows how to solve the equation, but has a problem with an alternative approach, and that is what the question is about. I would expect that a >80K user reads the whole question before posting, and answered accordingly
$endgroup$
– Martin R
Dec 3 '18 at 10:43
$begingroup$
Fair enough. Mistakes do happen.
$endgroup$
– DavidG
Dec 3 '18 at 10:47
|
show 1 more comment
$begingroup$
HINT
As noticed we can't use differentiation to obtain the result indeed in general
$$f'(x)=g'(x) not Rightarrow f(x)=g(x)$$
consider for example the simple case
$$2x+3=1 to 2=0$$
I suggest to use tangent half-angle identities by $t = tan frac x2$ to obtain
$$sqrt{2}sec x+tan x=1 iff sqrt{2}frac{1+t^2}{1-t^2}+frac{2t}{1-t^2}=1$$
answered Dec 3 '18 at 10:20
gimusigimusi
92.8k84494
$endgroup$
HINT
As noticed we can't use differentiation to obtain the result indeed in general
$$f'(x)=g'(x) not Rightarrow f(x)=g(x)$$
consider for example the simple case
$$2x+3=1 to 2=0$$
I suggest to use tangent half-angle identities by $t = tan frac x2$ to obtain
$$sqrt{2}sec x+tan x=1 iff sqrt{2}frac{1+t^2}{1-t^2}+frac{2t}{1-t^2}=1$$
answered Dec 3 '18 at 10:20
gimusigimusi
92.8k84494
edited Dec 3 '18 at 10:43
answered Dec 3 '18 at 10:20
gimusigimusi
92.8k84494
answered Dec 3 '18 at 10:20
gimusigimusi
92.8k84494
answered Dec 3 '18 at 10:20
gimusigimusi
92.8k84494
92.8k84494
3
$begingroup$
Is this an answer to the question “Can someone give a proper explanation to my attempt ?” ?
$endgroup$
– Martin R
Dec 3 '18 at 10:39
$begingroup$
@MartinR Yes you are right, I add some comment on that.
$endgroup$
– gimusi
Dec 3 '18 at 10:40
$begingroup$
No, but it’s a hint as labelled by ‘HINT’
$endgroup$
– DavidG
Dec 3 '18 at 10:41
2
$begingroup$
@DavidG: OP explicitly states that (s)he knows how to solve the equation, but has a problem with an alternative approach, and that is what the question is about. I would expect that a >80K user reads the whole question before posting, and answered accordingly
$endgroup$
– Martin R
Dec 3 '18 at 10:43
$begingroup$
Fair enough. Mistakes do happen.
$endgroup$
– DavidG
Dec 3 '18 at 10:47
|
show 1 more comment
3
$begingroup$
Is this an answer to the question “Can someone give a proper explanation to my attempt ?” ?
$endgroup$
– Martin R
Dec 3 '18 at 10:39
$begingroup$
@MartinR Yes you are right, I add some comment on that.
$endgroup$
– gimusi
Dec 3 '18 at 10:40
$begingroup$
No, but it’s a hint as labelled by ‘HINT’
$endgroup$
– DavidG
Dec 3 '18 at 10:41
2
$begingroup$
@DavidG: OP explicitly states that (s)he knows how to solve the equation, but has a problem with an alternative approach, and that is what the question is about. I would expect that a >80K user reads the whole question before posting, and answered accordingly
$endgroup$
– Martin R
Dec 3 '18 at 10:43
$begingroup$
Fair enough. Mistakes do happen.
$endgroup$
– DavidG
Dec 3 '18 at 10:47
3
3
$begingroup$
Is this an answer to the question “Can someone give a proper explanation to my attempt ?” ?
$endgroup$
– Martin R
Dec 3 '18 at 10:39
$begingroup$
Is this an answer to the question “Can someone give a proper explanation to my attempt ?” ?
$endgroup$
– Martin R
Dec 3 '18 at 10:39
$begingroup$
@MartinR Yes you are right, I add some comment on that.
$endgroup$
– gimusi
Dec 3 '18 at 10:40
$begingroup$
@MartinR Yes you are right, I add some comment on that.
$endgroup$
– gimusi
Dec 3 '18 at 10:40
$begingroup$
No, but it’s a hint as labelled by ‘HINT’
$endgroup$
– DavidG
Dec 3 '18 at 10:41
$begingroup$
No, but it’s a hint as labelled by ‘HINT’
$endgroup$
– DavidG
Dec 3 '18 at 10:41
2
2
$begingroup$
@DavidG: OP explicitly states that (s)he knows how to solve the equation, but has a problem with an alternative approach, and that is what the question is about. I would expect that a >80K user reads the whole question before posting, and answered accordingly
$endgroup$
– Martin R
Dec 3 '18 at 10:43
$begingroup$
@DavidG: OP explicitly states that (s)he knows how to solve the equation, but has a problem with an alternative approach, and that is what the question is about. I would expect that a >80K user reads the whole question before posting, and answered accordingly
$endgroup$
– Martin R
Dec 3 '18 at 10:43
$begingroup$
Fair enough. Mistakes do happen.
$endgroup$
– DavidG
Dec 3 '18 at 10:47
$begingroup$
Fair enough. Mistakes do happen.
$endgroup$
– DavidG
Dec 3 '18 at 10:47
|
show 1 more comment
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023863%2fsolve-sqrt2-sec-x-tan-x-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
6
$begingroup$
This is not an answer to your question, but I'd note that the differentiation step is not valid. Try to apply it to something like $x^2 + 1 = 2$. Differentiating should give $2x = 0$, so $x = 0$ has to be the solution. Clearly this isn't correct. The problem here is that you are trying to solve for a specific value of $x$; differentiating the two sides indicates that you are treating the quantities as functions, rather than the numbers they are. I am not sure why it seems to work out in this case.
$endgroup$
– platty
Dec 3 '18 at 10:12
$begingroup$
Please let me know if my answer with an alternative method could be useful, otherwise I'll delete that.
$endgroup$
– gimusi
Dec 3 '18 at 10:49
2
$begingroup$
Ah, I think I see why it happens to work out. It turns out that at $2 pi n - pi/4$, the equation $sqrt{2} tan x + sec x = 0$ is actually true. So in a way, you end up using the fact that your answer works to derive it, in a sort of circular argument. Of course, like I mentioned above, this method is invalid, but now we know why it appears to work.
$endgroup$
– platty
Dec 3 '18 at 10:59
2
$begingroup$
@platty thanx very much. you could post it as answer though !
$endgroup$
– ss1729
Dec 3 '18 at 13:33