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For what value of parameter $a$ the equation

$$ \text{arccos}(\sin(x)+a)=\text{arcsin}(\cos(x)+a) $$

has exactly one solution in the interval for $x$ $(0; 2\pi)$?

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    For $a=-\cos x_0-\sin x_0$ the equation is true only when $x=x_0$ – 2017-01-31
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    @polfosol How did you find this? – 2017-01-31
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    Expand $$\sin\left(\text{arccos}(\sin(x)+a)-\text{arcsin}(\cos(x)+a)\right)$$ – 2017-01-31
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    @polfosol For $x_0=\pi/2$ we have $a=-1$. But there is no solution in this case for $x=\pi/2$ – 2017-01-31
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    @polfosol For $a$ negative I suppose the equation has no solutions.. – 2017-01-31
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    Yes, you are right – 2017-01-31
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52791/discussion-between-widawensen-and-polfosol). – 2017-01-31

1 Answers 1

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Suppose that the equation has only one solution, say $x_0$. Let $$\alpha:=a+\,\sin x_0\\ \beta:=a+\cos x_0$$ then after taking sine and cosine from the both sides of your equation: $$\sin(\arccos{\alpha})=\beta\\ \cos(\arcsin{\beta})=\alpha$$ So we can write: $$\begin{align}\alpha^2+\beta^2&=\sin^2(\arccos{\alpha})+\cos^2(\arcsin{\beta}) \\ &=1-\alpha^2+1-\beta^2\\ \implies \alpha^2+\beta^2&=1 \end{align}$$ On the other hand: $$\begin{align}\alpha^2+\beta^2 &=(a+\sin x_0)^2+(a+\cos x_0)^2\\ &=1+2a^2+2a(\sin x_0+\cos x_0) \end{align}$$ Therefore $$1+2a^2+2a(\sin x_0+\cos x_0)=1\\ \implies a=-\sin x_0-\cos x_0\quad\text{or}\quad a=0$$ Now you can see that for $a=0$ every $x$ in $(0,\frac{\pi}2)$ satisfies the equation. So it is safe to say that if the equation has only one solution in $x_0$ then we must have $$a=-\sin x_0-\cos x_0=\sqrt{2}\cos(x_0+3\pi/4)$$ Of course this is a necessary and not sufficient condition. Taking this into account, I plotted the following figure:

plot

As you see, if $x_0\in[0,\,2\pi]$ satisfies your equation then we are sure that $\pi\le x_0\le \frac{3\pi}2$.

Since $a=\sqrt{2}\cos(x_0+3\pi/4)$ we can say that $a$ is a function of $x_0$ whose domain is $[\pi,\frac{3\pi}{2}]$. Also it is easy to verify that the range of this function is $[1,\,\sqrt{2}]$. So in conclusion:

  1. If $0\sqrt{2}$ then your equation has no solution in $[0,\;2\pi]$.
  2. If $1\le a\le\sqrt{2}$ then the solution of your equation is equivalent to the solution of $$\cos(x+3\pi/4)=\frac a{\sqrt 2}$$ which has two solutions for $a<\sqrt 2$ and only one for $a=\sqrt 2$.
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    It's very important result... so we have necessary condition. – 2017-01-31
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    I've made the graph for $f(x)=\arccos \left(\sin \left(x\right)+a\right)-\arcsin \left(\cos \left(x\right)+a\right)$. It seems that the correct answer is only $a=\sqrt{2}$. For other values $f(x)$ has two zeros or none. What do you think about it? It is very close to your solution.. – 2017-01-31
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    @Widawensen You can check the fact that for $a=1$ there is only one solution as well. And I didn't get what you mean by _why it is not visible in the graph of..._ – 2017-01-31
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    For a=1 there are two solutions on my graph $x_0=\pi$ and $x_0={3/2}\pi$ – 2017-01-31
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    @Widawensen that doesn't add up with $a=-\cos x_0-\sin x_0$ which was a _necessary_ condition – 2017-01-31
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    $arccos(sin(\pi)+1)=arcacos(1)=0, arcsin(cos(\pi)+1)=arcsin(0)=0, .... arccos(sin(3\pi/2)+1)=arccos(0)=\pi/2, arcsin(cos(3\pi/2)+1)=arcsin(1)=\pi/2$ – 2017-01-31
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    Additional checking $-cos(\pi)-sin(\pi)=1-0=1=a, -cos({3\pi/2})-sin({3\pi/2})=0+1=1=a$ – 2017-01-31
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    @Widawensen I plotted [another figure](https://i.stack.imgur.com/fev63.png). Looks like you were right all along. Let me give it some more thought... – 2017-01-31
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    I think Pol. that the graph you have made in your answer still is giving necessary conditions not sufficient.. – 2017-01-31
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    That is the kind of question I _can_ answer `:)` $$\,$$ Simply open another tab for asking a new question and then insert the image from your computer. The browser will upload it to the cloud and inserts the link into your question box. Copy that link and don't post that dummy question! @Widawensen – 2017-01-31
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    Pol. see my picture for the case https://i.stack.imgur.com/VElkv.jpg – 2017-01-31
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    @Polsofol I think that your solution is almost complete except the final step. Here we should have $\cos(x_0+3\pi/4)=1$ (or -1 but this we can somehow exclude) because only this gives unique solution for $x_0$ i.e. $x_0+3\pi/4= 2\pi$ and $ x_0=5\pi/4$ what is compliant with reality.. What do you think about it ? – 2017-01-31
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    @Widawensen Thanks. I edited the answer. Hope it is now OK – 2017-01-31
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    Thank you Polfosol for the answer for this, as we have seen, a little hard to interpret question :) – 2017-01-31
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    @Widawensen It was a pleasure. See you around – 2017-01-31