Question

How do you solve $2{\cos }^{2}x-3\cos x+1=0$ and find all solutions in the interval $(0,2\pi )$?

Answer 1

Determine all the value of $\mathit{x}$ which lies between $\mathbf{0}$ and $\mathbf{2}\mathbf{\pi }$:

$$\begin{gathered} 2{\cos }^{2}x-3\cos x+1=0 \\ \Rightarrow 2{\cos }^{2}x-2\cos x-\cos x+1=0 \\ \Rightarrow 2\cos x\left(\cos x-1\right)-1\left(\cos x-1\right)=0 \\ \Rightarrow \left(2\cos x-1\right)\left(\cos x-1\right)=0 \end{gathered}$$

Equate each factor to zero:

$\begin{array}{rcl}2\cos x-1& =& 0\\ & \Rightarrow & \cos x=\frac{1}{2}\\ & \Rightarrow & x=\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\end{array}$

And,

$$\begin{gathered} \cos x-1=0 \\ \Rightarrow \cos x=1 \\ \Rightarrow x=0 \end{gathered}$$

Neglect $0$ because the value of $x$ lie in the interval $0<x<2\mathrm{\pi }$.

Hence, the value of $\mathit{x}$ which lies between $\mathbf{0}$ and $\mathbf{2}\mathbf{\pi }$ is $\frac{\mathbf{\pi }}{\mathbf{3}}$ and $\frac{\mathbf{5}\mathbf{\pi }}{\mathbf{3}}$.