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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 $x=2n\pi-\frac{\pi}{4}$. But, what if I do the following: $$ \sqrt{2}\sec x+\tan x=1\\ \text{Differentiating}\implies\sqrt{2}\sec x\tan x+\sec^2 x=0\implies\sqrt{2}\tan x+\sec x=0 $$ Step 1 $$ \sec x=-\sqrt{2}\tan x=\frac{1-\tan x}{\sqrt{2}}\implies2\tan x=\tan x-1\implies\tan x=-1\\ \boxed{x=n\pi-\frac{\pi}{4}} $$ Step 2 $$ \tan x=1-\sqrt{2}\sec x=\frac{-\sec x}{\sqrt{2}}\implies2\sec x-\sqrt{2}=\sec x\\ \implies\sec x=\sqrt{2}\implies\boxed{x=2n\pi\pm\frac{\pi}{4}} $$ $$ x=n\pi-\frac{\pi}{4}\quad\&\quad x=2n\pi\pm\frac{\pi}{4}\\\implies \bigg[x=2n\pi-\frac{\pi}{4}\text{ or }x=2n\pi+\frac{3\pi}{4}\bigg]\quad\&\quad x=2n\pi\pm\frac{\pi}{4}\\ \implies \boxed{x=2n\pi-\frac{\pi}{4}} $$ In my attempt why do I need Step 2 to get the complete solution ?

Can someone give a proper explanation to my attempt ?

Sooraj S
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    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. – platty Dec 03 '18 at 10:12
  • Please let me know if my answer with an alternative method could be useful, otherwise I'll delete that. – user Dec 03 '18 at 10:49
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    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. – platty Dec 03 '18 at 10:59
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    @platty thanx very much. you could post it as answer though ! – Sooraj S Dec 03 '18 at 13:33

2 Answers2

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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, $\sqrt{2} \tan\left(2n\pi - \frac{\pi}{4} \right) + \sec\left(2n\pi - \frac{\pi}{4} \right) = 0$, 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.

platty
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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

$$\sqrt{2}\sec x+\tan x=1 \iff \sqrt{2}\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}=1$$

user
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    Is this an answer to the question “Can someone give a proper explanation to my attempt ?” ? – Martin R Dec 03 '18 at 10:39
  • @MartinR Yes you are right, I add some comment on that. – user Dec 03 '18 at 10:40
  • No, but it’s a hint as labelled by ‘HINT’ –  Dec 03 '18 at 10:41
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    @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 – Martin R Dec 03 '18 at 10:43
  • Fair enough. Mistakes do happen. –  Dec 03 '18 at 10:47
  • @MartinR I read the question and by comments the mistake has been already pointed out, therefore I've suggested an alternative effective method to solve that, since I suppose the asker was looking for that. I'll ask if it can be useful or not. – user Dec 03 '18 at 10:48