<s>[INST] 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?
 [/INST] Using auxiliary angle, we first change the equation
$$
\sin x+\cos x=\sin x \cos x \tag*{(*)} 
$$
equivalently to
$$
\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)= \frac{\sin 2 x}{2} 
$$
Putting $y=x-\dfrac{\pi}{4} $ gives a quadratic equation in $\cos y $
$$
\sqrt{2} \cos y=\frac{1}{2}\left(2 \cos ^{2} y-1\right)
$$
Solving the quadratic equation yields
$$
\cos y=\frac{\sqrt{2}-2}{2} \text { or } \frac{\sqrt{2}+2}{2} \text { (rejected) }
$$
Plugging  the general solution for $(*)$ is
$$
x=\frac{(8 n+1) \pi}{4} \pm \cos ^{-1}\left(\frac{\sqrt{2}-2}{2}\right),
$$
where $n\in Z.$
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