Equazione goniometrica formule parametriche

$\sin x + \cos x = \sqrt{2}$

Posto $t=\tan \frac{x}{2}$

$\frac{2t}{t^2+1}+\frac{1-t^2}{t^2+1}=\sqrt{2}$

$2t+1-t^2=\sqrt{2}t^2 + \sqrt{2}$

$(\sqrt{2}+1)t^2-2t+\sqrt{2}-1=0$

$t=  \frac{2 \pm \sqrt{4-4(\sqrt{2}+1)(\sqrt{2}-1)}}{2(\sqrt{2}+1)}$

$t=  \frac{2 \pm \sqrt{4(1-(\sqrt{2}+1)(\sqrt{2}-1))}}{2(\sqrt{2}+1)}$

$t=  \frac{2 \pm 2\sqrt{1-(\sqrt{2}+1)(\sqrt{2}-1)}}{2(\sqrt{2}+1)}$

$t=  \frac{1\pm \sqrt{1-(\sqrt{2}+1)(\sqrt{2}-1)}}{\sqrt{2}+1}$

$t=\frac{1 \pm \sqrt{1-(2-1)}}{\sqrt{2}+1}$

$t= \frac{1 \pm 0}{\sqrt{2}+1}$

$t=\sqrt{2}-1$

Per le identità degli angoli doppi: $\tan 2\theta = \frac{2 \tan \theta}{1- \tan^2 \theta}$, quindi:

$\tan x = \frac{2(\sqrt{2}-1)}{1-(\sqrt{2}-1)^2}= \frac{2(\sqrt{2}-1)}{2\sqrt{2}-2}= \frac{2(\sqrt{2}-1)}{2(\sqrt{2}-1)}=1$

$x=\arctan 1 = \frac{\pi}{4} +2\pi k$