Chapter 10.4, Problem 18E

Solution

To Solve: $3\cos x=\cos x-1$for $0\le x<2\pi$

  $x=\frac{5\pi }{6}\text{ or }x=\frac{7\pi }{6}$for $0\le x<2\pi$ .

Given information:

The equation given is $3\cos x=\cos x-1$

Concept and Formula Used:

To solve the above expression, use the values of inverse trigonometric ratios on the unit circle. Also, cosine function is negative in the second and third quadrant, so two solutions will be $\pi -x,\pi +x$ .

  $\begin{array}{l}\left(i\right){\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{\pi }{6}\\ \left(ii\right){\cos }^{-1}\left(\cos x\right)=x\end{array}$

Calculation:

The given equation is $3\cos x=\cos x-1$

Subtracting $\cos x$ from both the sides:

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

Dividing both the sides by 2

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

Taking inverse of cosine function both the sides

  $\begin{array}{c}{\cos }^{-1}\left(\cos x\right)={\cos }^{-1}\left(-\frac{1}{2}\right)\\ \theta ={\cos }^{-1}\left(-\frac{1}{2}\right)\\ \theta =\pi -\frac{\pi }{6}=\frac{5\pi }{6},\pi +\frac{\pi }{6}=\frac{7\pi }{6}\end{array}$

Conclusion:

  $x=\frac{5\pi }{6}\text{ or }x=\frac{7\pi }{6}$for $0\le x<2\pi$ .