Solve $\sin(x)\cos(x)=\sin(x)+\cos(x)$ My initial idea was
$$(\sin(x)\cos(x))^2=1+2\sin(x)\cos(x)$$
Let $t=\sin(x)\cos(x)$;
$$t^2=1+2t \quad\Leftrightarrow\quad t=1-\sqrt2$$
(since $1+\sqrt2>1$).
I.e.
$$\sin(x)\cos(x)=1-\sqrt2 \quad\Leftrightarrow\quad \tfrac12\sin(2x)=1-\sqrt2 \quad\Leftrightarrow\quad x=\tfrac{1}{2}\arcsin(2(1-\sqrt2))$$
but I didn't get any ‘elegant’ final solution. Any better ideas?
 A: HINT
You can also proceed as follows:
\begin{align*}
\sin(x)\cos(x) = \sin(x) + \cos(x) & \Longleftrightarrow 2\sin(x)\cos(x) = 2(\sin(x) + \cos(x))\\\\
& \Longleftrightarrow 1 + 2\sin(x)\cos(x) = 1 + 2(\sin(x) + \cos(x))\\\\
& \Longleftrightarrow (\sin(x) + \cos(x))^{2} = 1 + 2(\sin(x) + \cos(x))
\end{align*}
Then make the change of variable $t = \sin(x) + \cos(x)$.
Can you take it from here?
A: Consider $$f(x)=\sin(x)\cos(x)-\sin(x)-\cos(x)$$ To avoid squaring, use the tangent half-angle formula $x=2 \tan ^{-1}(t)$ and you need to solve for $t$ the quartic
$$t^4-4t^3-1=0$$ which shows two real solutions
$$t_\pm=1+\frac{1}{\sqrt{2}}\pm\sqrt{\frac{1}{2} \left(5+4 \sqrt{2}\right)}$$ which is not better but does not contain any falso root due to squaring.
A: Let $y=\dfrac{\sin x+\cos x}{\sqrt{1^2+1^2}}=\cos\left(x-\dfrac\pi4\right)$
$\implies -1\le y\le1$
$y^2=\dfrac{1+2\sin x\cos x}2$
We have $$\dfrac{y^2-1}2=\sqrt2y\implies y^2-2\sqrt2y-1=0$$
$$y= \sqrt2\pm\sqrt3$$
$$\implies y=\sqrt2-\sqrt3$$
