Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) = \dfrac{\sin^2(2x)}{4}$$
$$\sin^2(2x) - 4 \sin(2x) -4 = 0$$
Here we can use the notation $t = \sin(2x)$ with the condition that $t \in [-1,1]$.
$$t^2-4t-4=0$$
Solving this quadratic equation we get the solutions:
$$t_1 = 2+ 2\sqrt{2} \hspace{3cm} t_2 = 2 - 2\sqrt{2}$$
I managed to prove that $t_1 \notin [-1, 1]$ and that $t_2 \in [-1, 1]$. So the only solution is $t_2 = 2 - \sqrt{2}$. So we have:
$$\sin(2x) = 2 - 2\sqrt{2}$$
From this, we get:
$$2x = \arcsin(2-2\sqrt{2}) + 2 k \pi \hspace{3cm} 2x = \pi - \arcsin(2-2\sqrt{2}) + 2 k \pi$$
$$x = \dfrac{1}{2} \arcsin(2-2\sqrt{2}) + k \pi \hspace{3cm} x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin(2 - 2\sqrt{2})  + k \pi$$
Is this solution correct? It's such an ungly answer, that I kind of feel like it can't be right. Did I do something wrong?

As your error has been pointed out, I am providing a different way to tackle the problem without introducing extra solutions. 
 From the given equation, we have $1=(1-\sin x)(1-\cos x)$, which is equivalent to
$$1=\Biggl(1-\cos\left(\frac{\pi}2-x\right)\Biggr)\left(2\sin^2 \frac{x}{2}\right)=4\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\sin^2\frac{x}{2}$$
That is 
$$2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\sin\frac{x}{2}=\pm 1.\tag{1}$$
This means
$$\cos\left(\frac{\pi}{4}-x\right)-\cos\frac{\pi}{4}=\pm 1.$$
Therefore
$$\cos\left(\frac{\pi}{4}-x\right)=\frac{1\pm\sqrt{2}}{\sqrt{2}}.$$
But $\frac{1+\sqrt2}{\sqrt2}>1$, so
$$\cos\left(\frac{\pi}{4}-x\right)=\frac{1-\sqrt 2}{\sqrt2}.\tag{2}$$
Therefore
$$2n\pi+\left(\frac{\pi}{4}-x\right) =  \pm \arccos \frac{1-\sqrt 2}{\sqrt2}$$
for some integer $n$.  This gives us
$$x=\left(2n+\frac14\right)\pi \pm \arccos \frac{1-\sqrt 2}{\sqrt2}.\tag{3}$$
In fact there are also complex solutions to $(1)$, and they are given by
$$x=\left(2n+\frac{1}{4}\right)\pi\pm i\operatorname{arccosh} \frac{1+\sqrt 2}{\sqrt2}.\tag{4}$$
Note that $$\operatorname{arccosh} \frac{1+\sqrt 2}{\sqrt2}=\ln\left(\frac{1+\sqrt2+\sqrt{1+2\sqrt2}}{\sqrt2}\right).$$
All real and complex solutions to the original equation are given by $(3)$ and $(4)$.
Note that
$$\frac\pi4 + \arccos \frac{1-\sqrt 2}{\sqrt2}=\frac12\arcsin(2-2\sqrt2)+\pi$$
and
$$\frac\pi4 -\arccos \frac{1-\sqrt 2}{\sqrt2}=\frac{\pi}{2}-\frac12\arcsin(2-2\sqrt2)-\pi.$$
So your solutions only work for odd $k$.  Even values of $k$ do not give solutions.
