The part showing $ g ( x ) = g ( - x ) $ is correct. For more details, note that from the equation $ g ( x ) ^ 2 = g ( x ) g ( - x ) $, which is correctly obtained by you, you can get $ g ( - x ) ^ 2 = g ( x ) g ( - x ) $ by substituting $ - x $ for $ x $. Then adding the two equations and using simple algebra, you get $ \big( g ( x ) - g ( - x ) \big) ^ 2 = 0 $, which shows that $ g ( x ) - g ( - x ) = 0 $, or equivalently $ g ( x ) = g ( - x ) $.
The part that you conclude $ g ( x ) = x ^ n $ for an even $ n $ or $ g ( x ) = \cos x $ is definitely wrong. The only remarkable properties of $ g $ that you've managed to prove up to that moment are $ g ( 0 ) \in \{ 0 , 1 \} $ and $ g ( - x ) = g ( x ) $. There are infinitely many functions having those properties, including constant functions $ g ( x ) = 0 $ and $ g ( x ) = 1 $, $ g ( x ) = | x | $, $ g ( x ) = 1 - | x | $, $ g ( x ) = \begin{cases} 1 & x = 0 \\ \frac { \tan ( x ^ 2 ) } { x ^ 2 } & x \ne 0 \end {cases} $, $ g ( x ) = \begin{cases} 0 & x = 0 \\ x ^ 4 \sin \frac 1 { x ^ 2 } & x \ne 0 \end {cases} $, $ g ( x ) = \cosh ( a x ) = \frac { \exp ( a x ) + \exp ( - a x ) } 2 = \frac 1 2 ( A ^ x + A ^ { - x } ) $ for any constant $ a $ with $ A = e ^ a $ (or equivalently, any positive constant $ A $ with $ a = \log A $), and many many more examples. Of course, many of these functions don't satisfy the original functional equation, or the other one that you've obtained correctly: $ g ( 2 x ) + g ( 0 ) = 2 g ( x ) ^ 2 $. But then again the case $ g ( x ) = x ^ { 2 n } $ proposed by yourself doesn't satisfy those equations, which made me wonder how you managed to get to the conclusion of including it as a possible case. More importantly, the functions $ g ( x ) = 0 $ and $ g ( x ) = \cosh ( a x ) $ are solutions to the original functional equation, and you haven't even considered them as possible cases.
The part that you are seemingly trying to rule out the case $ g ( x ) = x ^ n $ is again certainly wrong (of course the conclusion is right, but your argument is not valid). The fact that $ g ( 0 ) $ can't be equal to $ 1 $ doesn't help you, as up to that point you only know that $ g ( 0 ) $ must be either $ 0 $ or $ 1 $, and $ g ( 0 ) = 0 ^ n \in \{ 0 , 1 \} $ in that case. At that point, you don't know whether both cases $ g ( 0 ) = 0 $ and $ g ( 0 ) = 1 $ must be true for some cases. Even if you knew that, the only thing you could achieve would be that the functions $ g ( x ) = x ^ n $ do not form the whole class of the solutions. I suppose you've mixed up "inclusive or" with "exclusive or" here, along with other mistakes. I'm inclined to add that it's usual to take $ 0 ^ 0 = 1 $, and as $ 0 $ is an even integer, in the special case of $ n = 0 $, $ g ( 0 ) = 1 $ would happen, as opposed to your claim. Indeed in that case we would have $ g ( x ) = x ^ 0 = 1 $ for every $ x $, and that gives a solution to the original functional equation.
The part saying "it should be $ \cos x $ with additional constant $ a $ or $ b $ as you would like. So the final form should be like this $ g ( x ) = b + \cos x $" is very vague. I edited your question and fixed several formatting problems and grammatical errors, but couldn't figure out your intention in that part, and left it almost unchanged. Even the constant $ a $ doesn't appear in the expression for $ g ( x ) $. I suppose you intended to write something like $ b + a \cos x $, $ b + \cos ( a x ) $ or $ b + \cos ( a + x ) $. If that is so and you had something of the form $ g ( x ) = c \cos ( a x + d ) + b $ in mind, I want to note that from $ g ( 0 ) \in \{ 0 , 1 \} $ you could find out some restrictions on the constants $ a $, $ b $, $ c $ and $ d $. In particular, you could see that the case $ g ( x ) = b + \cos x $ could not happen unless $ b \in \{ - 1 , 0 \} $. Of course, more restrictions could be obtained by using other properties of $ g $, most importantly the original equation (which for example rules out the case $ b = - 1 $ in the last observation). I added this remark to note that the same observation you (incorrectly) used to rule out the case $ g ( x ) = x ^ n $ could (correctly) be used here to get valid information. On the bright side, it seems that you somehow found out that indeed $ g ( x ) = \cos ( a x ) $ is a solution to the functional equation, and you could change your former statement which counted $ \cos x $ as a possibility, but for example ruled out $ g ( x ) = \cos ( 2 x ) $.
As I think many that are yet in the process of learning mathematics may find these educationally useful, I add some further remarks. The first one is that letting $ y = 0 $ in the original equation you could get $ g ( x ) = g ( 0 ) g ( x ) $. You did check the case $ x = y = 0 $, and did let $ y = x $ and $ y = - x $, but somehow missed that one, which is as simple an observation as the other ones, if not simpler. This would show you that if you had $ g ( 0 ) = 0 $, then $ g ( x ) $ would be constantly equal to zero, which is in fact a solution. This could help you to rule out $ g ( x ) = x ^ n $ in a glance, this time correctly.
At last, I want to note that so far we've listed three class of solutions for d'Alembert's functional equation:
- $ g ( x ) = 0 $;
- $ g ( x ) = \cos ( a x ) $ for some constant $ a $;
- $ g ( x ) = \cosh ( a x ) $ for some constant $ a $;
(the case $ g ( x ) = 1 $ is included both in the second and third class, both when $ a = 0 $). I suppose these all fall into the category of "actual" solutions you were seeking, as they are given in terms of very familiar special functions. The fact is that these are the only continuous solutions, and there are noncontinuous solutions, too, which have very wild behaviors. See for example d'Alembert functional equation: $ f ( x + y ) + f ( x - y ) = 2 f ( x ) f ( y ) $ for the classification of the solutions, and links given in Overview of basic facts about Cauchy functional equation for finding out more about the mentioned wild cases.
