<s>[INST] The following equation to solve :$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$ The following equation to solve :
$$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$$
My try:
$$\frac{2}{\sin 2x}=\sqrt{2}(\cos x+\sin x)$$
$$\left(\frac{2}{\sin 2x}\right)^2=(\sqrt{2}(\cos x+\sin x))^2$$
$$\left(\frac{2}{\sin 2x}\right)^2=2(1+\sin 2x)$$
$$2\sin^2 2x +2\sin ^3 2x=4$$
$$2\sin^2 2x +2\sin ^3 2x-4=0$$
$t=\sin 2x$
$$2t^3+2t^2-4=(t-1)(t^2+2t+4)$$
$$\sin 2x =1\\$$
is it right ?
 [/INST] Avoid squaring whenever possible as it immediately introduces extraneous roots.
Method $\#1:$ Put $u=\dfrac{\cos x+\sin x}{\sqrt2}\implies2u^2=1+\sin2x$
$$\implies\dfrac2{2u^2-1}=2u\iff2u^3-u-1=0$$
Clearly, $u=1$ is a root. Please check for the other roots.
$\implies1=\dfrac{\cos x+\sin x}{\sqrt2}=\cos\left(x-\dfrac\pi4\right)$
$\implies x-\dfrac\pi4=2m\pi$ where $m$ is any integer.
Method $\#2:$
$\cos x+\sin x=\sqrt2\cos\left(x-\dfrac\pi4\right)$
$\sin2x=\cos2\left(x-\dfrac\pi4\right)=\left(\sqrt2\cos\left(x-\dfrac\pi4\right)\right)^2-1$
</s>