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# Solve the equation $\arccos x + \arccos(x^2)=0$

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I want to ask help in understanding if my point of view is correct: I must solve the equation $\arccos x + \arccos(x^2)=0$.

The domain of $\arccos x + \arccos(x^2)$ is $-1 \leq x \leq 1$, and it is

$$\arccos x + \arccos(x^2)=0 \iff \arccos x = - \arccos (x^2)$$

Taking cosine both sides and using the fact that cosine is even (so it is $\cos (-\arccos (x^2))=\cos(\arccos(x^2))=x^2$), I get

$$x=x^2 \iff x=0 \vee x=1$$

But $x=0$ isn't a solution of the equation; why am I getting it? I think that it has something to do with the step from $\arccos x = - \arccos (x^2)$ to $x=x^2$, but I'm not sure why this happens. I think that it is because $a=b \implies \cos a= \cos b$ but the converse isn't true (because I would get $a=b +2k\pi \vee a=-b+2k\pi$), so I only get the implication

$$\cos (-\arccos (x^2))=\cos(\arccos(x^2)) \implies x=x^2 \iff x=0 \vee x=1$$

Instead of the equivalence $\iff$, so I only know that the set $\{0,1\}$ is a subset of the solution of $\{x \in [-1,1] \ \text{s.t.} \ \arccos x +\arccos(x^2)=0\}$; so I must check if all the solutions I get are valid or not. Is this correct? If yes, how can I add conditions to get an equivalence $\iff$ in the step I've mentioned? Thank you.

Apr 24, 2021
edited by Hitago  Apr 24, 2021
edited by Hitago  Apr 24, 2021

#1
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I have not worked through all your logic but I will respond to one bit.

"But  x=0  isn't a solution of the equation; why am I getting it? "

in order for acos(x) to be a function, its range must be limited. By convention this range will be $$0\le acosx\le \pi$$

and for this range,  x=0 is not a solution.

However,

Say you set the range to be   $$-\pi \le x\le 0$$  that would be ok, it would still be a function.

Now x=0 is a solution and x=1 is not.

So maybe that will help you understand why you are getting it. Here is the graph:

Red is     acos(x)

Blue is     acos(x^2)

And doted green is acos(x) if you change the range. Apr 24, 2021
#2
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Thanks for your answer, I have a doubt: why for $x=0$ it is $\arccos x$ not a function? I know that I must restrict the image of the cosine, conventionally, to $[0,\pi]$ to get an injective function so I can invert it, but it doesn't seem to me that $x=0$ creates a problem...indeed $\arccos(0)=\frac{\pi}{2}$. What am I missing? :D

Hitago  Apr 24, 2021
#3
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If you use the conventional limits of    0<=arccos <= pi

then you are right,

$$arccos(0)=\frac{\pi}{2}$$

However if you use the different limits of    -pi<=arccos<= 0

then

arccos(o)= -pi/2

Say the question was

$$cos(\theta)=0$$

$$\theta=acos(0)$$  but you have to extend this to

$$\theta=2\pi n\pm acos(0)\\\theta=2\pi n\pm \frac{\pi}{2} \qquad n\in Z$$

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Or if you look at it from the point of view of the x^2, you get the same type of outcome

if  x^2 = 9     x could equal  +3 or -3  both answers are valid.

BUT

sqrt(9)=+3      This is by convention,  the convention could just as easily say the answer is -3.

When dealing with some problems, these conventions mean extra checks are necessary.

Melody  Apr 25, 2021