Solve $\sin x + \cos x = \sin x \cos x.$

Note that when you square the equation

$$(\sin x + \cos x)^2 = (\sin x \cos x)^2$$

which can be factorized as

$$(\sin x + \cos x - \sin x \cos x)(\sin x + \cos x + \sin x \cos x)=0$$

you effectively introduced another equation $\sin x + \cos x =- \sin x \cos x$ in the process beside the original one $\sin x + \cos x = \sin x \cos x$. The solutions obtained include those for the extra equation as well.

Normally, you should plug the solutions into the original equation to check and exclude those that belong to the other equation. However, given the complexity of the solutions, it may not be straightforward to do so. Therefore, the preferred approach is to avoid the square operation.

Here is one such approach. Rewrite the equation $\sin x + \cos x = \sin x \cos x$ as

$$\sqrt2 \cos(x-\frac\pi4 ) = \frac12 \sin 2x = \frac12 \cos (2x-\frac\pi2 ) $$

Use the identity $\cos 2t = 2\cos^2 t -1$ on the RHS to get the quadratic equation below

$$\sqrt2 \cos(x-\frac\pi4) = \cos^2 (x-\frac\pi4 ) -\frac12$$

or

$$\left( \cos(x-\frac\pi4) - \frac{\sqrt2-2}2\right)\left( \cos(x-\frac\pi4) - \frac{\sqrt2+2}2\right)=0$$

Only the first factor yields real roots

$$x = 2n\pi + \frac\pi4 \pm \cos^{-1}\frac{\sqrt2-2}2$$

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.