Does $\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \implies \cos{\theta}=1$?

$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta}}}}...=1$$ $$(\cos{\theta})^{1/2+1/4+1/8+...}=1$$ $$(\cos{\theta})^{1}=1$$ $$\cos{\theta}=1$$ $$\theta=2k\pi,k\in\mathbb Z$$

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 $a=b$.

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 $$\theta=2k\pi,k\in\mathbb Z$$