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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?

mf67
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    As far as I can tell, your working out is correct, except for there being infinitely many solutions: $x = \frac{1}{2}(\arcsin(2(1-\sqrt{2})) + 2k\pi), \quad k \in \mathbb{Z}$ – Daniel P Mar 06 '22 at 14:51
  • WolframAlpha also can't find a simpler form – Daniel P Mar 06 '22 at 14:52
  • "since $1+\sqrt2>1$", should be "since $1+\sqrt2>0.5$" –  Mar 06 '22 at 14:54
  • RamanujanXV. Yes, thank you. – mf67 Mar 06 '22 at 14:55
  • One problem. E.g. $x_0=-0.488147$ is not a solution. The error comes from the quadration. How to eliminate these? – mf67 Mar 06 '22 at 15:18
  • The problem can be solved by using $t=\sin(x)+\cos(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$ and then solve for $x.$ – Reinhard Meier Mar 06 '22 at 15:29
  • Reinhard Meier. Could you please show? – mf67 Mar 06 '22 at 15:31
  • @mf67 $t=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$ results in $x=\arcsin(t/\sqrt{2})+2\pi n -\frac{\pi}{4}$ or $x=(2n+1)\pi - \arcsin(t/\sqrt{2}) -\frac{\pi}{4}$. – Reinhard Meier Mar 06 '22 at 15:38
  • In general, an equivalence like the last one in the question must not only have a "${}+2k\pi$" term to account for the fact that $\sin(y)=\sin(y+2k\pi),$ it must also include an alternative sequence of solutions due to the fact that $\sin(y) = \sin(\pi-y).$ In this particular problem, half of the "solutions" in each of these sequences are correct and the other half come from the loss of sign information when squaring. – David K Mar 06 '22 at 17:44

3 Answers3

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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?

Átila Correia
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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.

Claude Leibovici
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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 $$\dfrac{y^2-1}2=\sqrt2y\implies y^2-2\sqrt2y-1=0$$

$$y= \sqrt2\pm\sqrt3$$

$$\implies y=\sqrt2-\sqrt3$$

lab bhattacharjee
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