I can't just put random values. I need a systematic method. And How can u gaurentee f(x)>2√2 in [0.5,1]?

@Sid Your comment prompted me to look more closely at the naive answer I offered before. Let me know if this one is useful.

I told in the problem description, no calculator allowed... It's a test qstn... Sorry for the inconvenience.

@Sid $\quad$ I do understand that no-calculators-allowed would rule out finding the parametric limits with WolframAlpha. It would be nice to identify the algorithm used by WolframAlpha. However, if you look at the portions with $\space\pi\space$ in the limit, you can mentally, or at least with paper and pencil, try values of $\space n\space$ and see that the $\space\pi\space$ portions cancel conveniently and yield a fraction you can calculate manually.

see but, just because the function = $2\sqrt2$ when x = 1/4, 1 (by trying out values) doesn't mean those are the only values between [0,2]... we need to prove that those two are in fact the only two values... because for example, the value $f(3\pi/4)$ is attained thrice in the range... look at the graphical solution the other person provided... (which is also not allowed cuz u would need a calculator.), so its not obvious that $f(x) = 2\sqrt2$ has only two solution in the range $x \in [0,2]$

@Sid There are not two solutions; there are two ranges of values for $\space 0\le x \le 2.\quad $ The value ranges are $x\in\big\{[0,\space 0.25],\space [1,\space 1.25]\big\}\quad $ Please look at the answer again and you will see this.

yep and how are u sure these are the only 2 value ranges

and no the value ranges are $x \in [0.25, 0.5] \cup [1,1.25]$

@Sid I neglected the point $2n+1$ in my earlier limits because it is taken care of by $2n-1\;$. WolframAlpha also show. $$f(x)=\big(3-\sin(2\pi x)\big)\sin\bigg(\pi x - \frac{\pi}{4}\bigg)- \sin\bigg(3\pi x +\frac{\pi}{4}\bigg)\lt 0\\ \implies \quad \frac{2 \big(π n - \frac{3 π}{8}\big)}{\pi}<x <\frac{2 (\pi n + \frac{\pi}{8})}{\pi}$$ $$f(x)=\big(3-\sin(2\pi x)\big)\sin\bigg(\pi x - \frac{\pi}{4}\bigg)- \sin\bigg(3\pi x +\frac{\pi}{4}\bigg) >2\sqrt{2}\\ \implies \frac{4 n + 1}{2}<x<2 n + 1$$ There are no other regions so the original limits are the only ones for which $f(x)\in I$.

this has nothing to do with what im asking. my question boils down to: can $(sinx-cosx)(2+sinxcosx) \leq 2$ be solved in a more intuitive way that what the have provided. Thats all. no calculator, no wolfram, nothing. Please help in this regard if possible.

the result is 3:1... the guy above also confirms it...@Raffaele

@Sid I saw my error and corrected it but my original answer provided a picture of why things work in an intuitive fashion because all solutions could be easily seen in terms of $45^\circ$ angles where the known sin is $\frac{\sqrt{2}}{2}.\quad$ This answer is intuitive in that the limits can be found mentally or on paper given any natural number $n$.

@Sid Forget the original answer if you will. For this one, a couple of comments ago, I showed that all other regions of $"x"$ make $f(x)\notin I\quad$. The limits shown can be calculated with pencil and paper or even mentally. And remember. You are asking for help. You cannot make demands and expect a kind reply.

U are not allowed to use Wolfram alpha or any calculator for that matter! Look, if you know the answer post it, if u don't let it be, but please stop spamming, it's annoying.

And it can't be done mentally. Look at that graph above. It increases, decreases and increases. In the region f(3/4)< f(x) < max(f(x)) it attains the same value 4 times in the interval (0.25,1.25). It's not obvious as just putting values of x and claiming those are the only points it's equal to 2√2

@Sid You are not trying. Put any integer value of $n$ into the limits and you can calculate a range of $x$-values mentally. Any $x$ value in that range works. And, by seeing that there are $100$ of the $\frac14$ ranges, you can see that $25\%$ of the values $0\le x \le 100$ are valid members of $I$.

Ok but how do u get those limits without a calculator. That's the qstn in hand here

@Sid I mentioned in a comment that I did not know the algorithm used by WolframAlpha. That would be a good topic for another question. If you could find it, algebraically or via limits or calculus, you would have both parts of your answer: How to define the limits and how to calculate the limits both without a calculator. I cannot help you any more with this but let me know if you "do" post the question.

Uhm finding those limits was my doubt. I already checked Wolfram etc and also graphed the given equation. I asked it here thinking I will get a solution (without online calculator).

Anyways, thanks for the help, I'll wait for someone to solve the qstn