Solve the equation on the interval $\; 0 \le \theta \lt 2\pi $ Hey I have two questions for math exchange! Let me list them first and show you what I have tried. By the way one can not use a calculator on the test review!

Solve the equations on the interval $\; 0 \le \theta \lt 2\pi $: 
  
  
*
  
*$2 \cos^2\theta - 3\cos \theta + 1 = 0 $    
  
*$\cot \theta = 2 \cos \theta$
  


So for the first one I tried:
\begin{gather}
2 \cos^2\theta - 3\cos \theta + 1 = 0\\
(2\cos^2\theta - 1) \times (3\cos\theta - 1) = 0\\
\begin{aligned}
2 \cos^2 \theta -1 &= 0 &\text{ or }&&  3\cos\theta -1 =0\\
\cos^2\theta &= \dfrac{1}{2} &\text{ or }&&  \theta = \cos^{-1}\left(\dfrac{1}{3}\right)
\end{aligned}
\end{gather}
After that I just got confused... I can't use a calculator so I do not know what to do next..
So The next problem was even more diffcult:
$$\cot \theta = 2 \cos \theta$$
And then immediately after that I got confused since there are two theta's, and I have absolutely no clue what to do next!
On the anwser key for test review, the correct anwsers are:


*

*$$2 \cos^2\theta - 3\cos \theta + 1 = 0$$
Answer:$\qquad\left\{ 0,\;\dfrac{\pi}{3} ,\; \dfrac{5\pi}{3} \right\}$ 

*$$\cot \theta = 2 \cos \theta$$ 
Answer : $\qquad\left\{\dfrac{\pi}{6} ,\; \dfrac{\pi}{2},\; \dfrac{5\pi}{6},\;\dfrac{3\pi}{2} \right\}$ 


Help would be greatly appreciated. Thankyou for reading!

First question: let's call $x=\cos\theta$. Then you have a second order equation $2x^2-3x+1=0$. then you would get two solutions in general, $x_1$ and $x_2$ and the final solution will be $\theta=\arccos(x_1)+n\pi$ and $\theta=\arccos(x_2)+n\pi,\;n\in\mathbb{N}$
Second question, use the definition of $\cot\theta=\cos\theta/\sin\theta$. Then, the equation reads
$$\cos\theta\left(\frac{1}{\sin \theta}-2\right)=0$$
Then or $\cos\theta=0$, so $\theta=\pi/2+n\pi,\;n\in\mathbb{N}$ or $\frac{1}{\sin \theta}-2$, which leads to $\sin\theta=1/2$, or equivalently, $\theta=\pi/6+n\pi,5\pi/6+n\pi,\;n\in\mathbb{N}$
