Chapter 10.4, Problem 9E

Solution

To find: the general solution of the given equation.

  $x=\frac{\pi }{6}+2n\pi$ or $x=\frac{5\pi }{6}+2n\pi$ , where $n$ is any integer.

Given:

  $2\sin x-1=0$

Concept used:

The sine function is positive in I and II quadrants.

The sine function is periodic with a period of $2\pi$ .

Calculation:

Consider,

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

Now, one solution of $\sin x=\frac{1}{2}$ in the interval $0\le x<2\pi$ is,

  $\begin{array}{c}x={\sin }^{-1}\left(\frac{1}{2}\right)\\ =\frac{\pi }{6}\end{array}$

The other solution in the interval is,

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

Now since the sine function is periodic with a period of $2\pi$ , therefore the general solution is,

  $x=\frac{\pi }{6}+2n\pi$ or $x=\frac{5\pi }{6}+2n\pi$ , where $n$ is any integer.