Chapter 5, Problem 59RE

To determine

To find: the solutions of the given equation.

Answer to Problem 59RE

  $\begin{array}{l}x=\frac{7\pi }{12}+n\pi \hfill \\ x=\frac{11\pi }{12}+n\pi \hfill \end{array}$

Explanation of Solution

Given:

A trigonometric equation as $2\sin 2x+1=0$ .

Calculation:

Let us consider the equation solve as shown:

  $\begin{array}{l}2\sin 2x+1=0\\ \sin 2x=-\frac{1}{2}\end{array}$

Because sin x has a period of 2p, so find the solution in $\left[0,2\pi \right).$

These solutions are $2x=\frac{7\pi }{6}and\text{ 2}x=\frac{11\pi }{6}$

Finally add multiples of $2\pi$ to the obtained solutions in $\left[0,2\pi \right)$ to get the general form of solution.

  $\begin{array}{l}\begin{array}{l}2x=\frac{7\pi }{6}+2n\pi \\ x=\frac{7\pi }{12}+n\pi \end{array}\hfill \\ \begin{array}{l}2x=\frac{11\pi }{6}+2n\pi \\ x=\frac{11\pi }{12}+n\pi \end{array}\hfill \end{array}$

Where n is an integer.