<s>[INST] 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!
 [/INST] *

*$2\cos^2\theta-3\cos\theta+1=0$:


Use the change of variables $x=\cos\theta$, in order to write the equation as
$$
2x^2-3x+1=0
$$
Solving for $x$ yields 
$$
x=1\quad or\quad x=\frac{1}{2}
$$
so in terms of $\theta$:
$$
\theta=0\quad or \quad \theta=\frac{\pi}{3}\quad or \quad \theta=-\frac{\pi}{3}
$$


*$\frac{\cos\theta}{\sin\theta}=2\cos\theta$:


Multiplying both terms by $\sin\theta$ yields
$$
\cos\theta=2\sin\theta\cos\theta\quad \Leftrightarrow\quad \sin(\frac{\pi}{2}-\theta)=\sin2\theta
$$
And solving for $\theta$ gives us
$$
\frac{\pi}{2}-\theta = 2\theta +2n\pi \quad or \quad \frac{\pi}{2}-\theta = \pi- 2\theta+2n\pi,
$$
that is:
$$
\theta\in \{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\}
$$
</s>