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I was solving an equation which states,

$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1$$

I solved like this:

The given equation can be written as following,

$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...= \sqrt{1 \sqrt{1 \sqrt{1}}}...$$

$$\implies \cos{\theta}=1$$

$$\implies \theta=\arccos {1}$$

So solution is $2n\pi, n \in \mathbb Z$.

Have I solved it wrong way?

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  • $\begingroup$ I see no particular reason for it to be true. After all, $x^2 = 1^2$ doesn't imply $x = 1$. $\endgroup$ – Patrick Stevens Feb 21 '16 at 10:08
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    $\begingroup$ Take for example, $y=\sqrt{2\sqrt{2\sqrt{2\cdots}}}$ $\endgroup$ – GoodDeeds Feb 21 '16 at 10:09
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    $\begingroup$ $\sqrt{a\sqrt{a\cdots }}=a^{1/2+1/4+\cdots}=a$ and $\sqrt{b\sqrt{b\cdots }}=b$, so the implication is true. $\endgroup$ – user236182 Feb 21 '16 at 10:17
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    $\begingroup$ If anything the ellipses should be inside the square roots: $\sqrt{b \sqrt{b \sqrt{b\ldots}}}$. But "infinite expressions" are not expressions, and undefined unless some specific meaning is given to them using some appropriate limit (which of course must also converge). $\endgroup$ – Marc van Leeuwen Feb 23 '16 at 14:27
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$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1$$ $$(\cos{\theta})^{1/2+1/4+1/8+...}=1$$ $$(\cos{\theta})^{1}=1$$ $$\cos{\theta}=1$$ $$\theta=2k\pi,k\in\mathbb Z$$

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To answer the question in the title, let $A=\sqrt{a \sqrt{a \sqrt{a\cdots}}}$ and $B=\sqrt{b \sqrt{b \sqrt{b\cdots}}}$.

Then $A=B$ implies $aA=A^2=B^2=bB$ and so $a=b$.

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  • $\begingroup$ Why would $aA=A^2$? $\endgroup$ – Man_Of_Wisdom Feb 21 '16 at 14:53
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    $\begingroup$ @Man_Of_Wisdom, isn't $A=\lim a_n$ with $a_{n+1}=\sqrt{a\cdot a_n}$ and $a_0=1$? $\endgroup$ – lhf Feb 21 '16 at 15:38
  • $\begingroup$ See for instance math.stackexchange.com/questions/589288/…. $\endgroup$ – lhf Feb 21 '16 at 15:41
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Another answer

suppose that $$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1$$ square both side you get $$cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}...=1$$ We know that $$\ \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}...=1$$ thus $$\cos{\theta}*1=1$$ therefore $$\theta=2k\pi,k\in\mathbb Z$$

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Hint:

$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1.$$

Multiply by $\cos(\theta)$ and take the square root. You get

$$\sqrt{\cos(\theta)}=\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}}...=1.$$

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