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Let $\cos(x) = 2.$

$e^{ix} + e^{-ix} = 4;$

Put $e^{ix}=t.$

$$ t^2 - 4t + 1 = 0 \iff t = 2 + \sqrt3 \vee 2 - \sqrt3;$$

$ x = -i\log(2+\sqrt3)$ or $-i\log(2-\sqrt3);$ where $i^2 = -1.$

What is the significance of this value? Because there is no possible triangle where the base is twice the hypotenuse. Does this mean that $\cos^{-1}(x)$ is also a function like the Riemann Zeta function where we can extend the domain of the function on the argand plane?

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    "Does this mean that cos^-1(x) is also a function like the Riemann Zeta function where we can extend the domain of the function on the argand plane?" - yes. A lot of the elementary functions can be analytically continued. Frankly, I'm surprised that you know that $\zeta$ admits a continuation when you were caught by surprise that the more elementary functions also admit continuation. – 2017-01-07
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    The Taylor series for $\cos$ converges for all $z$, thus defining an entire function (i.e., a holomorphic function that is defined on all of $\Bbb C$) – 2017-01-07
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    Please look at how I've edited your question to use LaTeX, which makes the equations prettier (indeed, makes them readable at all!). It's very similar to what you typed, but looks WAY better. :) – 2017-01-07
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    Thanks. I'll try that next time. – 2017-01-07
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    So, is there a pole for this function? Like S=1 for the reimann zeta function. – 2017-01-07
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    @RaghavChaturvedi: no, the exponential has no pole at finite distance. – 2017-01-07
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    $\arccos$ has branch cuts instead of poles. – 2017-01-07
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    @J.M.isn'tamathematician I'm not very suprised that someone would know about the analytic continuation of $\zeta$ and not $\cos$, because of the wide exposure (through a popular YouTube video and other means) of the claim that $1+2+3+4+\cdots=-1/12$. – 2017-01-07
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    @David, that sounds awfully like trying to learn how to walk when one doesn't even know how to crawl. – 2017-01-07
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    @J.M.isn'tamathematician Yes, it does, and that is a criticism I think I recall seeing made against attempts to explain that particular claim to non-mathematicians. – 2017-01-07
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    @Raghav Chaturvedi: Nah, the two singularities are at +/-1 (real numbers) and are branch points and not poles, meaning that if you analytically continue here this one will actually extend back on itself to cease being a function and instead become a relation (that is not a function). – 2017-01-07

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When cosine is supplied with an imaginary number, it becomes a hyperbolic cosine:

$$\cos(ix) = \cosh(x)$$

No factor of $i$ on the right -- it is a straight-up, hyperbolic cosine.

Thus when you take $\cos(-i \log(2 + \sqrt{3}))$ to obtain 2, this is equivalent to taking $\cosh(-\log(2 + \sqrt{3}))$, which also equals $\cosh(\log(2 + \sqrt{3}))$ since $\cos$ and $\cosh$ are even.

The circular trigonometric functions $\cos(x)$ and $\sin(x)$ are usually interpreted in terms of a triangle, but the better way to interpret them is they parameterize the circle by arc length, that is, $(\cos(t), \sin(t))$ traverses the circle at constant speed as $t$ advances.

Now, if we are considering complex inputs, these coordinates become complex numbers and the resulting shape is now a sheet in 4 dimensions, not a curve in 2. Taking imaginary "angles" into the function takes us onto another part of this 4-dimensional figure which looks like a unit hyperbola, not a circle, and the interpretation of $\cosh(\log(2 + \sqrt{3}))$ geometrically is it is the $x$-coordinate of a point on the unit hyperbola when you have traveled distance $\log(2 + \sqrt{3})$ of arc length along it, and that moving this distance thus leaves you at an $x$-coordinate of 2. I found this figure on the web which illustrates it.

enter image description here

(Image public domain from: https://en.wikipedia.org/wiki/File:Hyperbolic_functions-2.svg)