Solve the following equation algebraically. Give a general solution.
3cos^{2}x+cosx=2
in the interval $\displaystyle [0, 2\pi]$ ...
$\displaystyle \cos{x} = {\color{red}+}\frac{2}{3}$ ... $\displaystyle x = \arccos\left(\frac{2}{3}\right)$ and $\displaystyle x = 2\pi - \arccos\left(\frac{2}{3}\right)$
$\displaystyle \cos{x} = -1$ ... you tell me what x value on the unit circle has a cosine of -1
if you want all solutions, add integer multiples of $\displaystyle 2\pi$ to all three solutions.
$\displaystyle \cos{x} = \frac{2}{3}$ has two solutions, one in quad I and one in quad IV
for $\displaystyle x = \pi$, the general solution should be written ...
$\displaystyle x = \pi + 2k\pi , k \in \mathbb{Z}$
... and put away the calculator.