Chapter 7.3, Problem 116E

To determine

To calculate:

Thevalue of $x$ for the given trigonometric equations.

Answer to Problem 116E

The general solution for the equation $6\cos x-2=1$ is $x=2n\pi \pm \frac{\pi }{3}$ , for any integer $n$ .

Explanation of Solution

Given information:

  $6\cos x-2=1$

Simplify the equations $6\cos x-2=1$ is as follows:

Add $2$ on both sides of the equation:

  $\begin{array}{c}6\cos x-2+2=1+2\\ 6\cos x=3\end{array}$

Divide each term by $6$ and simplify:

  $\begin{array}{c}6\cos x=3\\ \frac{6\cos x}{6}=\frac{3}{6}\\ \cos x=\frac{1}{2}\end{array}$

Take the inverse cosine on both sides of the above equation to find the value of $\theta$ .

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

The exact value of ${\cos }^{-1}\left(\frac{1}{2}\right)$ is $\frac{\pi }{3}$ .

The cosine function is positive in the first and fourth quadrants. To find the second solution subtract the reference angle from $2\pi$ to find the solution in the fourth quadrant.

  $\begin{array}{c}x=2\pi -\frac{\pi }{3}\\ =\frac{6\pi -\pi }{3}\\ =\frac{5\pi }{3}\end{array}$

The period of the $\cos x$ functions is $2\pi$ so values will repeat every $2\pi$ radians in both directions.

Hence, the required general solution for the equation $6\cos x-2=1$ is $x=2n\pi \pm \frac{\pi }{3}$ , for any integer $n$ .