Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$ Question:
$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $
Answer key:
$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $
I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:
$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $
$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies 
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too
 A: $f(x)$ has period $2$.
What happens in $[0,2]$ is repeated $50$ times in the same way, thus the ratio we have in this interval is the ratio of all the race.
Run points are $1/4$ and swim point are $3/4$, therefore the ratio $swim/run$ is $3$.
How did I find the points in the image below.
Solve
$$(\sin (2 \pi  x)-3) \sin \left(\frac{\pi }{4}-\pi  x\right)-\sin \left(3 \pi  x+\frac{\pi }{4}\right)=2 \sqrt{2}$$
solutions are $x=\frac12+2k;\;x=1+2k;\;k\in\mathbb{Z}$.
In the interval $[0,2]$ the are $\frac12,1$.
Solve
$$(\sin (2 \pi  x)-3) \sin \left(\frac{\pi }{4}-\pi  x\right)-\sin \left(3 \pi  x+\frac{\pi }{4}\right)=0$$
solutions are $x=\frac14+2k;\;\frac54+2k;\;k\in\mathbb{Z}$
In the interval $[0,2]$ the are $\frac14,\frac54$.
The set where $0\le f(x)\le 2\sqrt 2$ is $\left[\frac14,\frac12\right]\cup\left[1,\frac54\right]$
The length is $\frac12$ and represents $\frac14$ of the interval $[0,2]$ while the other part is $\frac34$. Thus the second part is three times the first part.

$$...$$

A: let $t=\sin \theta -\cos \theta $ then we get $$\sin \theta \cos \theta =\frac{1-t^2}{2}$$ indeed we have tofind $\theta$ which satisfies the inequality $$t(2+\frac{1-t^2}{2})\le 2$$ $$\iff f(t)=t^3-5t+4\ge 0$$ As you say this came in an exam lets presume that the factorisation is not so complicated.Suppose it has an integer root then by rational root theorem we can say that any integer root $p$ (say) must divide $4$ .$p|4$ so if there is an integer root it has to be one of  $\pm1,\pm2\pm3\pm4$ We can skip $\pm 2\pm 3\pm4$ as $|t|\le \sqrt{2}$ .We are left with $\pm 1$ .By inspection we note that $f(1)=0$ hence we have found one factor i.e $t-1$ .The other factor can be easily found by long division ie $t^2+t-4$.
Now its easy to see $t^2+t-4<0$ for all $|t|\le ]\sqrt{2}$.
So $$t^3-5t+4\ge 0\implies (t-1)(t^2+t-4)\ge 0\implies t\le 1$$
Which  you can  be easily taken from here
A: Hint
$$f(x)=(3-\sin (2 \pi  x)) \sin \left(\pi x -\frac{\pi }{4}\right)-\sin \left(3 \pi x+\frac{\pi }{4}\right)$$ Expand and manipulate the trigonometic functions to obtain
$$f(x)=\frac{7 \big[\sin (\pi  x)- \cos (\pi  x)\big]-\big[\sin (3 \pi  x)+\cos (3 \pi  x)\big]}{2\sqrt{2}}$$
A: $$(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$$
Let $x = \cos \theta$ and $y = \sin \theta$. Then we get
\begin{align}
   (y - x)(2+xy) &\leq 2 \\
   x^2 + y^2 &= 1
\end{align}
We can now argue
\begin{align}
      -(y-x)^2 &= -1 + 2xy \\
   5 - (y-x)^2 &=  2(2 + xy) \\
\hline
   (y-x)(5 - (y-x)^2) &\le 4
\end{align}
Let $z = y-x$ and we find
\begin{align}
   z(5 - z^2) &\le 4 \\
   z^3 - 5z + 4 \ge 0 \\
   (z-1)(z^2+z-4) \ge 0 \\
   \hline
   z \ge \dfrac{\sqrt{17}-1}{2} \\
   -\frac{\sqrt{17}+1}{2} \le z \le 1
\end{align}
The first answer is superfluous. The second needs to be modified to
$$-\sqrt{2} \le z \le 1$$
$$-\sqrt{2} \le \sin \theta - \cos \theta \le 1$$
$$-\sqrt{2} \le \sqrt 2 \sin\left(x - \frac{\pi}{4} \right) \le 1$$
$$-1 \le \sin\left(x - \frac{\pi}{4} \right) \le \frac{1}{\sqrt 2}$$
$$\theta \in \left[(2n-1)\pi, \frac{\pi}{2}(4n+1) \right] \qquad \forall n \in \Bbb Z$$
A: Let $$A:=\sin\theta-\cos\theta,\qquad B:=\sin\theta\cos\theta$$
Then, we have
$$A^2=(\sin\theta-\cos\theta)^2=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta=1-2B$$
To solve $$A(2+B)-2\le 0$$
let us see if $A(2+B)-2$ can be factorized.
To see if $A(2+B)-2$ can be factorized, let us add $k(-A^2-2B+1)$ (which equals $0$) to $A(2+B)-2$ where $k$ is a non-zero real number.
$$\begin{align}A(2+B) -2&=2A+AB-2
\\\\&=2A+AB-2+\underbrace{k(-A^2-2B+1)}_{=0}
\\\\&=k\bigg(-A^2+\frac 2kA+\frac{k-2}{k}+\frac Bk(A-2k)\bigg)\end{align}$$
Now we want
$$-A^2+\frac 2kA+\frac{k-2}{k}$$
to have $A-2k$ as a factor, so solving
$$-(2k)^2+\frac 2k(2k)+\frac{k-2}{k}=0$$
gives$$k=\frac 12,\frac{-1\pm\sqrt{17}}{4}$$
So, we see that $k=\frac 12$ works.
Therefore, we have
$$\begin{align}A(2+B)-2&=A(2+B)-2+\frac 12(-A^2-2B+1)
\\\\&=\frac 12\bigg(-A^2+4A-3+2B(A-1)\bigg)
\\\\&=\frac 12\bigg((A-1)(-A+3)+2B(A-1)\bigg)
\\\\&=\frac 12(-A+2B+3)(A-1)\end{align}$$
So, all we need is to solve
$$\frac 12(-A+2B+3)(A-1)\le 0$$
A: We will use
$$(\sin\theta- \cos\theta)^2 
= \sin^2\theta - 2\sin\theta \cos\theta + \cos^2\theta
= 1 - 2\sin\theta \cos\theta$$
a few times.
$$(\sin \theta -\cos \theta)(2+\sin \theta \cos \theta)- 2 \le 0 $$
$$2[(\sin \theta -\cos \theta)(2+\sin \theta \cos \theta)- 2] \le 0 $$
$$(\sin \theta -\cos \theta)(4+2\sin \theta \cos \theta)- 4 \le 0 $$
$$(\sin \theta -\cos \theta)(5-(1-2\sin \theta \cos \theta))- 4 \le 0 $$
$$(\sin \theta -\cos \theta)(5-(\sin\theta - \cos\theta)^2)- 4 \le 0 $$
$$-(\sin\theta - \cos\theta)^3 + 5(\sin\theta - \cos\theta)-4 \le 0$$
$$(\sin\theta - \cos\theta - 1)
(-(\sin\theta - \cos\theta)^2 - (\sin\theta - \cos\theta) + 4) \le 0$$
$$(\sin\theta - \cos\theta - 1)
(-1 + 2\sin\theta\cos\theta - \sin\theta + \cos\theta) + 4) \le 0$$
$$(\sin\theta - \cos\theta - 1)
(\cos\theta - \sin\theta + 2\sin\theta\cos\theta + 3) \le 0$$
A: Imagine radius vector rotation in the four quadrants
At $\theta$ (0,180) sine vanishes
At  $\theta$ (90,270 ) cos vanishes
So the four zones are determined and demarcated as four the quadrants conveniently.
$$F(\theta)=(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) - 2$$
For extremum values only center values of each quad $F(\theta)$ needs to be evaluated/examined that too only for its sign.
We need only to evaluate at odd central values
$$ (2k-1)\dfrac{\pi}{4}$$
First quadrant
$$ 0 (...) -2 \le 0 \text{ okay} $$
Second quadrant
$$(2/\sqrt2)(2-\frac12) >0 \text{ not okay}$$
Third quadrant
$$ (2\sqrt2)(2-1/\sqrt2) \le 0 \text { okay} $$
Fourth quadrant
$$(2 \sqrt2)(....) \le 0 \text { okay} $$
So the correct sign occurs in 1st, 2nd, 4th quadrants only.
$ (\pi/2<\theta<3 \pi/2 ) $ interval is inadmissible.
No need for  a polar graph to be drawn... even if drawn for confirmation it would not be single valued radius in the second quadrant anyways..
A: I have re-written this answer because, with the help of WolframAlpha here, I was surprised to discover
$$0\le \big(3-\sin(2\pi x)\big)\sin(\pi x - \frac{\pi}{4})-  \sin(3\pi x +\frac{\pi}{4})\le 2\sqrt{2}\\
\implies (2 n - 1)<x\le \frac{2 \big(\pi n - \frac{3 π}{8}\big)}{\pi} \land n \in\mathbb{ Z}\\
\land \implies \frac{2 \big(\pi n + \frac{\pi}{8}\big)}{\pi}\le x\le\frac{(4 n + 1)}{2} \land n\in\mathbb{Z} $$
What the first "implies" means for $\space 0\le x\le 100\space$ is that there are discreet ranges of values that make this true as in these samples for $1\le n\le 6$
$$(1\le x \le \frac{5}{4}=1.25)\quad 
(3\le x \le \frac{13}{4}=3.25)\quad 
(5\le x \le \frac{21}{4}=5.25)\quad \\
(7\le x \le \frac{29}{4}=7.25)\quad 
(9\le x \le \frac{37}{4}=9.25)\quad 
(11\le x \le \frac{45}{4}=11.25)\quad $$
What the second "implies" means is that the following ranges apply for $\space x$ for $1\le n\le 6$.
$$\frac{9}{4}=2.25\le x \le \frac{5}{2}=2.5\qquad 
\frac{17}{4}=4.25\le x \le \frac{9}{2}=4.5\\ 
\frac{25}{4}=6.25\le x \le \frac{13}{2}=6.5\qquad 
\frac{33}{4}=8.25\le x \le \frac{17}{2}=8.5\\ 
\frac{41}{4}=10.25\le x \le \frac{21}{2}=10.5\qquad 
\frac{49}{4}=12.25\le x \le \frac{25}{2}=12.5$$
We can see that each value of $n$ generates $\space x\space$ with a total "coverage of $\space 0.5\qquad $ Note that $\quad 1\le n \le 24\implies 1\le x \le 97\quad$ so there are $24\times0.5=12$ "full-numbers-worth" of $0\le x \le 100$ that satisfy $I.\quad$ However, the first "implies" above permits
$n=25\implies 99\le x \le 99.25\space $ and the second "implies" peermits  $n=0.0\implies 0.25\le x \le 0.5\quad $. There are a total of $100$ intervals of $0.25$ each so the members run $\frac14$ of the time.
This means that there members swim $\frac34$ of the time and the ratio of $\frac{swim}{run}$ is $\quad 3:1$
