Solving $\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x}$ Solve $\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x}$. 
My attempt: 
$\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x} \Rightarrow \cos(x-\alpha)\cos(x-\beta)-\cos{\alpha}\cos{\beta} = \sin^2{x},$ and
$$\begin{aligned}LHS & = \frac{1}{2}\cos\left(2x-\alpha-\beta\right)+\frac{1}{2}\cos\left(\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha+\beta\right)\\& = \frac{1}{2}\cos\left(2x-\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha+\beta\right) \\& = -\sin\left(\frac{2x-\alpha-\beta+\alpha+\beta}{2}\right)\sin\left(\frac{2x-\alpha-\beta-\alpha-\beta}{2}\right) \\& = -\sin{x}\sin\left(x-\alpha-\beta\right)\end{aligned}$$
Thus, we have: 
$$\begin{aligned} & -\sin{x}\sin\left(x-\alpha -\beta\right) = \sin^2{x} \\& \Rightarrow \sin^2{x}+\sin{x}\sin\left(x-\alpha-\beta\right) = 0 \\& \Rightarrow \sin{x}\left(\sin{x}+\sin\left(x-\alpha-\beta\right)\right) = 0\end{aligned} $$
So either $\sin{x} = 0$ or $\sin{x} = \sin\left(\alpha+\beta-x\right)$. If $\sin{x} = 0$, then we have $\sin{x} = 0 \Rightarrow \sin{x}$ $= \sin\left(0\right)$ $\Rightarrow x = n\pi$ or $(2n+1)\pi$ for $n\in\mathbb{Z}$ -- we can write this as $k\pi$, where $k\in\mathbb{Z}$. If, on the other hand, $\sin{x} = \sin\left(\alpha+\beta-x\right)$ then $x = 2n\pi+\alpha+\beta-x \Rightarrow x = n\pi+\frac{1}{2}\left(\alpha+\beta\right)$, or $ x = 2n\pi+\alpha+\beta-x $ (EDIT: this was meant to be $x = (2n+1)\pi-\alpha-\beta+x$), which contains no solutions (is that the right way to put it?). Thus the solutions for the equation are $n\pi+\frac{1}{2}\left(\alpha+\beta\right)$ and $k\pi$, for any integers $k$ and $n$. 
Question: The answer in the book has the condition $\alpha+\beta \ne (2m+1)\pi$ -- why is that?
Request: If you have the time, please critique the way my solution is presented (reasoning, use of notation, flow etc  - I've an admission test in which this plays big part soon). Is my use of $\Rightarrow$ ok?  
 A: One point: why are you going back from $x = n\pi+\frac{1}{2}(\alpha+\beta)$ to $x=2n\pi + \alpha+\beta-x$? And why do you say there are no solutions?
That said: you are incorrect in claiming that $\sin(x) = \sin(\alpha+\beta-x)$ implies that $x = 2n\pi + \alpha+\beta-x$. That is not the only ways in which two values of sine may be equal. After all, sine takes the same value at $0$ and at $\pi$, even though $0$ and $\pi$ don't differ by a multiple of $2\pi$. And it takes the value $\frac{1}{2}$ at both $\frac{\pi}{6}$ and at $\frac{5\pi}{6}$, even though the difference is not even an integer multiple of $\pi$. So that part of the derivation is incorrect.
Added/Edit. From the comments, it seems you meant to write the correct conditions: either $x$ and $\alpha+\beta-x$ differ by a multiple of $2n\pi$ (this is the condition you checked), or else they are "symmetric around $\pi/2$ or $3\pi/2$ up to a multiple of $2\pi$", which gives the condition $x = (2m+1)\pi - (\alpha+\beta-x)$.
Your error was discarding/completely ignoring the second condition. The second condition does not give an equation for $x$, but it does give an equation for $\alpha+\beta$! It says that you must have $0 = (2m+1)\pi - (\alpha+\beta)$, or that $\alpha+\beta=(2m+1)\pi$ must hold. In other word: if $\alpha+\beta = (2m+1)\pi$, then you will always have $\sin(x) = \sin(\alpha+\beta-x)$, regardless of what the value of $x$ is.
So, in summary: if $\alpha+\beta = (2m+1)\pi$ for some integer $m$, then every $x$ is a solution. And if $\alpha+\beta\neq (2m+1)\pi$, then your derivation is correct (once you fix the analysis of this second case).
