<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] There is a subtle way to get rid of the extra solutions that uses the work you've done.
When you render
$\sin 2x=2\sin x\cos x =2-2\sqrt{2}$
you then have with $u=\sin x, v=\cos x$:
$uv=1-\sqrt{2}$
$\color{blue}{u+v=uv=1-\sqrt{2}}$
where the blue equation reimposes the original requirement and it's goodbye extraneous roots.  Using the Vieta formulas for a quadratic polynomial this is solved by rendering $u$ and $v$ as the two roots of the quadratic equation
$w^2-(1-\sqrt2)w+(1-\sqrt2)=0$
The roots of this equation are obtained by the usual methods, and because of the symmetry of the original equation between $\sin x$ and $\cos x$ you may take either root as the sine and the other as the cosine.  Note that with the negative product the values must be oppositely signed which informs us of quadrant location.
$\sin x=\dfrac{1-\sqrt2+\sqrt{2\sqrt2-1}}{2},\cos x=\dfrac{1-\sqrt2-\sqrt{2\sqrt2-1}}{2}$ ($x$ in 2nd quadrant)
$\sin x=\dfrac{1-\sqrt2-\sqrt{2\sqrt2-1}}{2},\cos x=\dfrac{1-\sqrt2+\sqrt{2\sqrt2-1}}{2}$ ($x$ in 4th quadrant)
Then with the quadrant information above we may render the correct roots for $x$:
$x=\pi-\arcsin(\dfrac{1-\sqrt2+\sqrt{2\sqrt2-1}}{2})+2n\pi$
$x=\arcsin(\dfrac{1-\sqrt2-\sqrt{2\sqrt2-1}}{2})+2n\pi$
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