1. Let's solve the quadratic equation $x^{2}-5x+6=0$ first. 2. The standard form of a quadratic equation is $ax^{2}+bx+c=0$. Here, $a=1$, $b=-5$, and $c=6$. 3. We use the quadratic formula: $$x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ 4. Calculate the discriminant: $$\Delta = b^{2}-4ac = (-5)^{2} - 4 \times 1 \times 6 = 25 - 24 = 1$$ 5. Since $\Delta > 0$, there are two real roots. 6. Substitute values into the formula: $$x=\frac{-(-5) \pm \sqrt{1}}{2 \times 1} = \frac{5 \pm 1}{2}$$ 7. Calculate the two roots: - $x_1 = \frac{5 + 1}{2} = 3$ - $x_2 = \frac{5 - 1}{2} = 2$ 8. So, the solutions to the quadratic equation are $x=3$ and $x=2$. 9. Now, solve the trigonometric equation $2\cos(\theta) = 1$. 10. Divide both sides by 2: $$\cos(\theta) = \frac{1}{2}$$ 11. Recall that $\cos(\theta) = \frac{1}{2}$ at angles $\theta = \pm \frac{\pi}{3} + 2k\pi$, where $k$ is any integer. 12. Therefore, the general solutions are: $$\theta = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$$ Final answers: - Quadratic roots: $x=3$ and $x=2$ - Trigonometric solutions: $\theta = \frac{\pi}{3} + 2k\pi$ or $\theta = -\frac{\pi}{3} + 2k\pi$ for any integer $k$.