Chapter 5, Problem 47RE

Solution

To calculate: The values of $x$ for $2\cos x<1$ , where $0\le x<2\pi$ .

The solution set for the given inequality is $\left(\frac{\pi }{3},\frac{5\pi }{3}\right)$ .

Given information:

The given inequality is $2\cos x<1$ for $0\le x<2\pi$ .

Calculation:

First of all, replace the inequality sign with equality sign and simplify the equation $\sin 2x=2\cos x$ for $x$ .

Divide both sides of the given equation $\sin 2x=2\cos x$ for $x$ .

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

Since the periodicity of cosine is $2\pi$ , there are two solutions for the equation $\sin 2x=2\cos x$ in the given interval and the solutions are $x=\frac{\pi }{3}$ and $x=\frac{5\pi }{3}$ .

So, the solution set for the given inequality is $\left(\frac{\pi }{3},\frac{5\pi }{3}\right)$ .