$2\text{sin}^2∝-3\text{sin}∝-2=0$
$↔ 2\text{sin}^2∝-4\text{sin}∝+\text{sin}∝-2=0$
$↔ 2\text{sin}∝(\text{sin}∝-2)+(\text{sin}∝-2)=0$
$↔ (\text{sin}∝-2)(2\text{sin}∝+1)=0$
$↔ \left[ \begin{array}{l}\text{sin}∝=2\\\text{sin}∝=-\dfrac{1}{2}\end{array} \right.$
Loại $\text{sin}∝=2$ vì $-1≤\text{sin}∝≤1$
Với $\text{sin}∝=-\dfrac{1}{2}$, ta có:
$\text{cos}^2∝+\text{sin}^2∝=1$
$↔ \text{cos}^2∝=1-\Bigg(-\dfrac{1}{2}\Bigg)^2$
$↔ \text{cos}^2∝=\dfrac{3}{4}$
$→ \text{cos}∝=±\dfrac{\sqrt[]{3}}{2}$
$→ \text{tan}∝=\dfrac{\text{sin}∝}{\text{cos}∝}=±\dfrac{\sqrt[]{3}}{3}$
