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?