Chapter 7, Problem 37RE

To determine

To find:The solution of the equation $2{\cos }^{2}x-7\cos x+3=0$ in the interval $\left[0,2\pi \right)$ .

Answer to Problem 37RE

The solutions of the equation $2{\cos }^{2}x-7\cos x+3=0$ are $x=\frac{\pi }{3},\frac{5\pi }{3}$ .

Explanation of Solution

Given information:

The equation is $2{\cos }^{2}x-7\cos x+3=0$ and interval is $\left[0,2\pi \right)$ .

Calculation:

Simplify the equation.

  $\begin{array}{c}2{\cos }^{2}x-7\cos x+3=0\\ 2{\cos }^{2}x-6\cos x-\cos x+3=0\\ 2\cos x\left(\cos x-3\right)-\left(\cos x-3\right)=0\\ \left(\cos x-3\right)\left(2\cos x-1\right)=0\end{array}$

Simplify further.

  $\begin{array}{c}\cos x=3\text{or}\cos x=\frac{1}{2}\\ x={\cos }^{-1}\left(3\right)\text{or}x={\cos }^{-1}\left(\frac{1}{2}\right)\end{array}$

Since cosines function lies between $-1$ and $1$ . So, there are no solutions for the first equation $\cos x=3$ . In the interval $\left[0,2\pi \right)$ , the solutions for the second equation are $x=\frac{\pi }{3}$ and $x=\frac{5\pi }{3}$ .

Therefore, the solutions of the equation $2{\cos }^{2}x-7\cos x+3=0$ are $x=\frac{\pi }{3},\frac{5\pi }{3}$ .