[tex]\displaystyle \sf 2cos^2(x)-3cos(x)-2 = 0 \ \ ;\ \ x\in [0,\pi] \\\\ \text{Fatorando, some e subtraia cos(x) }: \\\\ 2cos^2(x)-3cos(x)-cos(x)+cos(x)-2= 0 \\\\ 2cos^2(x)-4cos(x)+cos(x)-2= 0 \\\\ 2cos(x)\cdot \left[cos(x)-2\right]+\left[cos(x)-2\right] = 0 \\\\ \left[cos(x)-2\right]\cdot \left[2cos(x)+1\right] = 0 \\\\ cos(x)-2=0\to cos(x)=2 \to \not\exists \ x\in\mathbb{R} }}\checkmark[/tex]

[tex]\displaystyle \sf 2cos(x)+1 = 0 \to cos(x) = \frac{-1}{2} \to\boxed{\sf x = \frac{2\pi }{3}} \\\\\\ \Large\boxed{\sf n^{o}\ \text{de Solu\c c\~oes em }[0,\pi] = 1 }[/tex]