[tex]5\sin^2 x-2\cos^2x-3\sin x\cos x=0\\
\text{ Impartind ecuatia la}\ \cos^2 \text{x se obtine:}\\
5 \cdot \dfrac{sin^2x}{cos^2 x}-2-3\cdot \dfrac{\sin x}{\cos x}=0\\
5 tg^2 x-3 tg\ x-2=0\\
\text{Notam:} tg\ x=t\\
5t^2-3t-2=0\\
\Delta= 9+40=49\Rightarrow \sqrt{\Delta}=7\\
t_1=\dfrac{3+7}{10}= 1\\
t_2=\dfrac{3-7}{10}= -\dfrac{2}{5}[/tex]
[tex]i)tg\ x=1 \\
S_1= \{ \dfrac{\pi}{4} +k\pi,k\in \mathbb{Z}\}\\
ii) tg\ x=-\dfrac{2}{5}\\
S_2= \{ -arctg\frac{2}{5}+k\pi ,k\in \mathbb{Z}\}\\
S=S_1 \cup S_2[/tex]
