Chapter 5, Problem 46RE

Solution

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

The solution set for the given inequality is $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ .

Given information:

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

Calculation:

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

Subtract $2\cos x$ from both sides of the equation $\sin 2x=2\cos x$ .

  $\sin 2x-2\cos x=0$

Apply the Double-Angle identity $\sin 2u=2\sin u\cos u$ into the obtained equation and simplify.

  $\begin{array}{c}2\sin x\cos x-2\cos x=0\\ 2\cos x\left(\sin x-1\right)=0\end{array}$

Now, it can be concluded that $2\cos x=0$ or $\sin x-1=0$ .

Simplify the equation $2\cos x=0$ for $x$ .

  $\begin{array}{c}\cos x=0\\ x={\cos }^{-1}0\\ x=\frac{\pi }{2}\end{array}$

Since the periodicity of cosine is $2\pi$ , there are two solutions for the equation $2\cos x=0$ in the given interval and the solutions are $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$ .

Simplify the equation $\sin x-1=0$ for $x$ .

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

Since the periodicity of sin is $\pi$ , there are two solutions for the inequality $\sin x-1>0$ in the given interval and the solutions are $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$ .

So, the solution set for the given inequality is $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ .