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Question:-

Consider the function

$$f(x)=\sin x+\cos x+\tan x+\cot x+\sec x+\csc x$$

Let $P$  be the smallest possible positive real number such that the equation $f(x)=P$ has real solutions, and let $N$ be the largest possible negative real number such that the equation $f(x)=N$  has real solutions.

Find the value of $[1000(P+N)]$.

MyApproach:

I am not getting idea how to proceed, since I can't understand how to find the real solutions. If I somehow find the smallest positive and largest negative from the real solutions, then it would be possible for me to solve this question.

Any idea how to get the real solutions from f(x)?

Community
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    What happened to the hint posted by @gimusi –  Feb 12 '18 at 08:14
  • "deleted by owner" – Blue Feb 12 '18 at 08:14
  • @Blue may I know why?Is it my fault that I have asked a bad question or his fault that he has answered incorrectly.And who is the owner of maths stackexchange? –  Feb 12 '18 at 08:20
  • hint: consider v=cos(x)+sin(x) and try to express the function in terms of v. – Ben Feb 12 '18 at 08:21
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    https://math.stackexchange.com/questions/769027/solve-sin-x-cos-x-tan-x-csc-x-sec-x-cot-x-2-in-the-interval-0x2-pi and https://math.stackexchange.com/questions/775269/if-sin-a-cos-a-tan-a-cot-a-sec-a-csc-a-7-then-x2-44x-3 – lab bhattacharjee Feb 12 '18 at 08:23
  • @labbhattacharjee great answer on the link that you mentioned ,I upvoted it.But here if I follow the same I would not be able to find the value of C and S since the value of $f(x)$ is not mentioned. –  Feb 12 '18 at 08:28
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    @CalculusProgrammer: Don't worry. Your question is fine. Everything is fine. "deleted by owner" only means "deleted by author". :) I guess gimusi simply decided the answer fell short somehow.. – Blue Feb 12 '18 at 08:31
  • @Blue OK thanks for telling that.By the way if you interested in my question you can answer it,may be I will get to learn something more since you have such a high reputation. –  Feb 12 '18 at 08:34
  • @lab The question you post has little to do with this question. Here we are searching for range, wheres there we were looking for roots for particular value $(2)$. Maybe I am wrong who knows. – King Tut Feb 12 '18 at 08:38
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    @KingTut, You are not wrong. I was trying to link related problems to the given expression. – lab bhattacharjee Feb 12 '18 at 08:40
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    @lab in that case you did good, methods from that question might apply here. – King Tut Feb 12 '18 at 08:42
  • No time to write up a solution right now, but solving symbolically via Maple, the values of $P,N$ are: $$P=2+3\sqrt{2}\approx 6.242640686$$ $$N=1-2\sqrt{2}\approx -1.828427124$$ – quasi Feb 12 '18 at 09:11

2 Answers2

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Hint:

Let $\sin x+\cos x=y\implies2\sin x\cos x=y^2-1$

$$a=\sin x+\cos x+\tan +\cot x+\sec x+\csc x=y+\dfrac2{y^2-1}+\dfrac{2y}{y^2-1}$$

$$\iff y^3-ay^2+y+a+2=0$$

Now $$-\sqrt2\le y\le\sqrt2\implies-2\le-y^2\le0$$

Case $\#1:$

If $a\le0,$

$$0=y^3-ay^2+y+a+2\le2\sqrt2-2a+\sqrt2+a+2$$ $$\iff a\le3\sqrt2+2,\text{ but }a\le0$$

and $$0=y^3-ay^2+y+a+2\ge-2\sqrt2-2a-\sqrt2+a+2$$ $$\iff a\ge2-3\sqrt2\implies 0\le a\le2-3\sqrt2$$

Case $\#2:$

If $a>0$

Please try yourself , I'm yet to complete it.

lab bhattacharjee
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Period of $f$ is $2\pi$ so lets analyse it in the length $[0,2\pi]$ with the help of $f'$. $$f'(x) = \cos x - \sin x+\sec^2x-\csc x \cot x+\sec x\tan x-\csc^2 x = 0\\ $$

Now you have to factorise this to get the roots.

$$\frac{ (\cos(x) - \sin(x)) ( \sin(x) \cos(x) - \sin(x)- \cos(x))}{2 (\cos(\tfrac{x}{2}) - \sin(\tfrac{x}{2}))^2\sin^2(\tfrac{x}{2}) }=0$$

Now after finding the roots of $f'$, which should not be many in number, its easier to manually check for value of $f$ at these points.

I get two trivial roots as $\frac{\pi}{4}, \frac{5\pi}{4}$. There are more roots from the factor $(\sin(x) \cos(x) - \sin(x)- \cos(x))$ which I havent evaluated.

King Tut
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