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Solve $$\cos x \sin x + |\cos x + \sin x| = 1.$$

My sister sent me this problem. I am helping her for it, but since I'm not exactly at my best mental state, I want someone to see if I get the result right. Thanks!

I start by defining

$$ y = |\cos x + \sin x| \Rightarrow y \geq 0 $$

and note that

$$ y^2 = \cos^2x + \sin^2x + 2\sin x \cos x \\ \Rightarrow y^2 = 1 + 2\sin x \cos x \\ \Rightarrow \cos x \sin x = \frac{y^2 - 1}{2} $$

Now I can rewrite the initial equation

$$ \cos x \sin x + |\cos x + \sin x| = 1 \\ \Rightarrow \frac{y^2 - 1}{2} + y = 1 \Rightarrow y^2 - 1 + 2y = 2 \Rightarrow y^2 + 2y - 3 = 0 $$

The last equation has two solutions: $y = 1$ or $y = -3$. Since $y \geq 0$, we take $y = 1$.

With $y = 1$, we have:

$$ |\cos x + \sin x | = 1 \\ \Rightarrow \left| \sqrt{2} \cos\left(x - \frac{\pi}{4}\right) \right| = 1 \\ \Rightarrow \left| \cos\left(x - \frac{\pi}{4}\right) \right| = \frac{1}{\sqrt{2}} \\ \Rightarrow \begin{cases} \cos\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \Rightarrow x - \frac{\pi}{4} = \pm \frac{\pi}{4} + 2\pi k \Rightarrow \begin{cases} x = \frac{\pi}{2} + 2\pi k \\ x = 2\pi k\end{cases} \\ \cos\left(x - \frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \Rightarrow x - \frac{\pi}{4} = \pm \frac{3\pi}{4} + 2\pi k \Rightarrow \begin{cases}x = \pi + 2\pi k \\ x = -\frac{\pi}{2} + 2\pi k\end{cases} \end{cases} \forall k \in \mathbb{Z} \\\\ \Rightarrow \boxed{ \begin{cases} x = \pm \frac{\pi}{2} + 2\pi k \\ x = k\pi \end{cases} \forall k \in \mathbb{Z} } $$

I want to know if my solution is right. What do you think?

Gary
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ducbadatcs
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    From https://math.stackexchange.com/tags/solution-verification/info: This tag should be used when you have a proposed solution to a problem and have specific concerns or doubts about the validity of that solution. A question with this tag should include an explanation for why the argument presented is not convincing enough. – Martin R Feb 05 '25 at 09:32
  • @MartinR I said I wasn't in my best mental state at the beginning of the question, and it can lead to a derailed solution. – ducbadatcs Feb 06 '25 at 10:42

2 Answers2

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Looks fine.

Alternatively, the equation can be written $$\eqalign{|\cos x+\sin x|=1-\cos x\sin x\ &\Leftrightarrow\ (\cos x+\sin x)^2=(1-\cos x\sin x)^2\cr &\Leftrightarrow\ 4\cos x\sin x=\cos^2x\sin^2x\cr & \Leftrightarrow\ \cos x\sin x=0\ ,\cr}$$ discarding $\cos x\sin x=4$ which has no real solution. Now this is easy to solve, $$x=\frac{n\pi}{2}\ ,$$ which is equivalent to your solution.

David
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Let us avoid squaring which immediately introduces extraneous roots which demand exclusion.

Actually for real $\cos\left(\dfrac\pi4-x\right),$

$$\left|\cos\left(\dfrac\pi4-x\right)\right|=\dfrac1{\sqrt2}\iff\cos^2\left(\dfrac\pi4-x\right)=\dfrac12$$

Method$\#1:$ Using $\cos2y=2\cos^2y-1,$

$$\cos2\left(\dfrac\pi4-x\right)=0\iff\sin2x=0$$

Can you take it from here?

Method$\#2:$

Using $\cos^2y=\cos^2A\iff\sin^2y=\sin^2A$

As Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, we have $y=n\pi\pm A$

$$\dfrac\pi4-x=m\pi\pm\dfrac\pi4\iff x=-m\pi\text{ or }x=\dfrac\pi2-m\pi$$

Now, can you please show the equivalence of the two results!

lab bhattacharjee
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  • Not sure why the first line of your answer says "Let us avoid squaring" and then the third line involves... squaring. When we can be certain that both sides of an equation are positive, squaring will never introduce extraneous roots. That is why in your third line you can write $\Leftrightarrow$ and not just $\Rightarrow$. – David Feb 05 '25 at 23:42
  • @David, Just wondering, is $\sin x\cos x=4$ not producing extraneous roots? – lab bhattacharjee Feb 06 '25 at 01:11
  • @David, Also, for real $a,b;$ $$ |a|=|b|\iff a^2=b^2$$ But $a=b$ does not necessarily imply $a^2=b^2$ as we are tacitly introducing $a+b=0$ – lab bhattacharjee Feb 06 '25 at 01:13
  • "Is $\sin x\cos x=4$ not producing extraneous roots?" - of course not because it has no real roots, $\sin x\cos x$ always lies between $-\frac12$ and $\frac12$ if $x$ is real. – David Feb 06 '25 at 01:22
  • "$|a|=|b|\Leftrightarrow a^2=b^2$" - true, that is an example of what I already said with "when we can be certain that both sides of an equation are positive". – David Feb 06 '25 at 01:23
  • "$a=b$ does not necessarily imply $a^2=b^2$" - yes it does. If you still believe otherwise then please give an example of $a,b$ such that $a=b$ and $a^2\ne b^2$. – David Feb 06 '25 at 01:25
  • @David, Sorry, I meant when we are squaring $a=b,$ we are immediately introducing $a+b=0$ – lab bhattacharjee Feb 06 '25 at 01:35
  • Also, extraneous roots may be imaginary as well, right? Can you please share some authentic link if you think otherwise. – lab bhattacharjee Feb 06 '25 at 01:37
  • I think it's pretty clear that in this case the OP is looking for real solutions, so there is no need to discuss imaginary roots. If you want to look for complex solutions then I suggest you post a new question. You will need to realise that $|z|^2=z^2$ is not always true for complex numbers. – David Feb 06 '25 at 01:45