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Solve for $x$ the equation $\arccos(2x)-\arccos(x)=\pi/3$.

My attempt using $\cos(a-b)$. But it gives me a sqrt-expression that I don't know how to handle.


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up vote 1 down vote accepted

This reduces to an easy quadratic. First, take the cosine of both sides. Using the cosine addition formula, I get

$$2 x \cdot x + \sqrt{1-4 x^2} \sqrt{1-x^2} = \frac12$$

Manipulate and square both sides to get

$$\left ( 2 x^2-\frac12\right)^2 = (1-4 x^2)(1-x^2) = 1-5 x^2+4 x^4$$


$$3 x^2=\frac{3}{4} \implies x = \pm \frac{1}{2}$$

Now plugging in both answers, one sees that only the $-1/2$ result makes sense for the principal branch (or any branch) of the arccos. Thus, $x=-1/2$.

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Why is it only x=-1/2 that satisfies the equation? I set up: acos(1) - acos(0.5)= pi/3 and acos(-1) - acos(-0.5)= pi/3 Now I can get multiple results, since acos(1)=pi/2 and also =3pi/4. How should I think about that? – jacob Sep 29 '13 at 11:58
@jacob: $\arccos{1}=0$, $\arccos{(1/2)}=\pi/3$. – Ron Gordon Sep 29 '13 at 12:24

Hint: $\arccos\left(2x\right)=\arccos\left(x\right)+\pi/3$. Now take the cosinus on both sides and see what happens.

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Avoid squaring which often incurs the burden of extraneous roots

Let $\displaystyle \arccos x=\theta\ \ \ \ (1)\implies x=\cos\theta$

We have $\displaystyle \arccos(2\cos\theta)-\theta=\frac\pi3\implies \arccos(2\cos\theta)=\theta+\frac\pi3\ \ \ \ (2) $

As the principal value of $\arccos$ lies $\in[0,\pi],$

from $(1), 0\le\theta\le\pi$

form $(2), 0\le\theta+\frac\pi3\le\pi\implies -\frac\pi3\le\theta\le\frac{2\pi}3$

So, we need $\displaystyle 0\le \theta\le \frac{2\pi}3\ \ \ \ (3)$

$\displaystyle (2)\implies 2\cos\theta=\cos\left(\frac\pi3+\theta\right)=\cos\frac\pi3\cos\theta-\sin\frac\pi3\sin\theta$

$\displaystyle \implies\frac{\sqrt3}2\sin\theta=\left(\frac12-2\right)\cos\theta$

$\displaystyle \implies\tan\theta=-\sqrt3=\tan\left(-\frac\pi3\right)$

$\displaystyle \implies \theta=n\pi-\frac\pi3$ where $n$ is any integer

From $\displaystyle(3), 0\le n\pi-\frac\pi3\le \frac{2\pi}3\iff 0\le3n-1\le2\implies n=1$

Can you take it from here?

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