Chapter 7.4, Problem 45E

To determine

To find:The solution of the equation $2{\cos }^2\theta -7\cos \theta +3=0$ .

Answer to Problem 45E

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

Explanation of Solution

Given information:

The given equation is $2{\cos }^2\theta -7\cos \theta +3=0$ .

Calculation:

Simplify the given equation.

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

First substitute the factor $2\cos \theta -1=0$ .

Simplify the equation

  $\begin{array}{c}2\cos \theta -1=0\\ 2\cos \theta =1\\ \cos \theta =\frac{1}{2}\end{array}$

As cosine has period $2\pi$ . So first find the solutions in $\left[0,2\pi \right)$ . The function cosine is positive in first and fourth quadrants.

Therefore, the solution to the equation in $\left[0,2\pi \right)$ are

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{1}{2}\right)\\ =\frac{\pi }{3},\frac{5\pi }{3}\end{array}$

As cosine repeats its values after every $2\pi$ units.

So, the general solution of the equation is given by

  $\theta =\frac{\pi }{3}+2k\pi ,\frac{5\pi }{3}+2k\pi$

Therefore, the solutions of the equation $2\cos \theta -1=0$ are $\theta =\frac{\pi }{3}+2k\pi$ and $\theta =\frac{5\pi }{3}+2k\pi$ .

Now substitute the factor $\cos \theta -3=0$ .

As the value of the function cosine is always between $-1$ to $1$ . Therefore, $\cos \theta -3=0$ is not possible for any value of $\theta$ .

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