$2\left(2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\right)^2 + 2\sin\dfrac{x}{2}\cos\dfrac{x}{2} + 2\cos^2\dfrac{x}{2} - 1 = 0$
$\Leftrightarrow 8\sin^2\dfrac{x}{2}\cos^2\dfrac{x}{2} + 2\sin\dfrac{x}{2}\cos\dfrac{x}{2} + 2\cos^2\dfrac{x}{2} - 1 = 0$
$\Leftrightarrow 8\sin^2\dfrac{x}{2} + 2\tan\dfrac{x}{2} + 2 - \dfrac{1}{\cos^2\dfrac{x}{2}} = 0$
$\Leftrightarrow 8\dfrac{\tan^2\dfrac{x}{2}}{\dfrac{1}{\cos^2\dfrac{x}{2}}} + 2\tan\dfrac{x}{2} + 2 - \left(1 + \tan^2\dfrac{x}{2}\right)= 0$
$\Leftrightarrow \dfrac{8\tan^2\dfrac{x}{2}}{1 + \tan^2\dfrac{x}{2}} + 2\tan\dfrac{x}{2} - \tan^2\dfrac{x}{2} + 1 = 0$
$\Leftrightarrow 8\tan^2\dfrac{x}{2} + 2\tan\dfrac{x}{2}\left(1 + \tan^2\dfrac{x}{2}\right) - \tan^2\dfrac{x}{2}\left(1 + \tan^2\dfrac{x}{2}\right) + 1 + \tan^2\dfrac{x}{2} = 0$
$\Leftrightarrow -\tan^4\dfrac{x}{2} + 2\tan^3\dfrac{x}{2} + 8\tan^2\dfrac{x}{2} + 2\tan\dfrac{x}{2} + 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\tan\dfrac{x}{2} \approx - 1,8663\\\tan\dfrac{x}{2} \approx 4,0901\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\dfrac{x}{2} \approx \arctan(- 1,8663) + k\pi\\\dfrac{x}{2} \approx \arctan4,0901 + k\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x \approx 2\arctan(- 1,8663) + k2\pi\\x \approx 2\arctan4,0901 + k2\pi\end{array}\right. \qquad (k \in \Bbb Z)$
