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!
 A: 
Solve $2\cos^2\theta - 3\cos\theta + 1 = 0$ in the interval $[0, 2\pi)$.

As learner pointed out in the comments, you factored incorrectly.  To split the linear term, we must find two numbers with product $2 \cdot 1 = 2$ and sum $-3$.  They are $-2$ and $-1$.  Hence,
\begin{align*}
2\cos^2\theta - 3\cos\theta + 1 & = 0\\
2\cos^2\theta - 2\cos\theta - \cos\theta + 1 & = 0\\
2\cos\theta(\cos\theta - 1) - 1(\cos\theta - 1) & = 0\\
(2\cos\theta - 1)(\cos\theta - 1) & = 0 
\end{align*}
Setting each factor equal to zero yields
\begin{align*}
2\cos\theta - 1 & = 0 & \cos\theta - 1 & = 0\\
2\cos\theta & = 1 & \cos\theta & = 1\\
\cos\theta & = \frac{1}{2} & \cos\theta & = \cos 0\\
\cos\theta & = \cos\left(\frac{\pi}{3}\right)
\end{align*}
Consider the diagram below.

By symmetry, if $\cos\theta = \cos(-\theta)$.  Since coterminal angles have the same cosine, 
$$\cos\theta = \cos\varphi  \implies \theta = \pm\varphi + 2n\pi, n \in \mathbb{Z}$$
Hence, 
$$\cos\theta = \cos\left(\frac{\pi}{3}\right) \implies \theta = \pm \frac{\pi}{3} + 2n\pi, n \in \mathbb{Z}$$
and 
$$\cos\theta = \cos 0 \implies \theta = 0 + 2n\pi, n \in \mathbb{Z}$$
We want those solutions in the interval $[0, 2\pi)$.
\begin{align*}  
0 & \leq \frac{\pi}{3} + 2n\pi < 2\pi \implies n = 0 \implies \theta = \frac{\pi}{3}\\
0 & \leq -\frac{\pi}{3} + 2n\pi < 2\pi \implies n = 1 \implies \theta = \frac{5\pi}{3}\\
0 & \leq \theta = 0 + 2n\pi < 2\pi \implies n = 0 \implies \theta = 0
\end{align*}
Hence, the solution set is $S = \left\{0, \dfrac{\pi}{3}, \dfrac{5\pi}{3}\right\}$.

Solve $\cot\theta = 2\cos\theta$ in the interval $[0, 2\pi)$.

\begin{align*}
\cot\theta & = 2\cos\theta\\
\frac{\cos\theta}{\sin\theta} & = 2\cos\theta\\
\cos\theta & = 2\sin\theta\cos\theta\\
0 & = 2\sin\theta\cos\theta - \cos\theta\\
0 & = \cos\theta(2\sin\theta - 1)
\end{align*}
Setting each factor equal to $0$ yields
\begin{align*}
\cos\theta & = 0 & 2\sin\theta - 1 & = 0\\
\cos\theta & = \cos\left(\frac{\pi}{2}\right) & 2\sin\theta & = 1\\
& & \sin\theta & = \frac{1}{2}\\
& & \sin\theta & = \sin\left(\frac{\pi}{6}\right)
\end{align*}
By the same reasoning as above, 
$$\cos\theta = \cos\left(\frac{\pi}{2}\right) \implies \theta = \pm \frac{\pi}{2} + 2n\pi$$
Consider the diagram above.  By symmetry, $\sin\theta = \sin(\pi - \theta)$.  Since coterminal angles have the same sine, 
$$\sin\theta = \sin\varphi \implies \theta = \varphi + 2n\pi, n \in \mathbb{Z}~\text{or}~\theta = \pi - \varphi + 2n\pi, n \in \mathbb{Z}$$
Hence, 
$$\sin\theta = \sin\left(\frac{\pi}{6}\right) \implies \theta = \frac{\pi}{6} + 2n\pi, n \in \mathbb{Z}~\text{or}~\theta = \pi - \frac{\pi}{6} + 2n\pi, n \in \mathbb{Z}$$
We wish to solve the equation in the interval $[0, 2\pi)$.  I leave it to you to determine the values of $n$ such that 
\begin{align*}
0 & \leq \frac{\pi}{2} + 2n\pi < 2\pi\\
0 & \leq -\frac{\pi}{2} + 2n\pi < 2\pi\\
0 & \leq \frac{\pi}{6} + 2n\pi < 2\pi\\
0 & \leq \pi - \frac{\pi}{6} + 2n\pi < 2\pi
\end{align*}
and the corresponding values of $\theta$.
A: 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, $\theta=\pi/6+n\pi,5\pi/6+n\pi,\;n\in\mathbb{N}$
A: *

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


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:
$$
\theta\in \{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\}
$$
