Chapter 10.4, Problem 22E

Solution

To describe: and correct the error in solving the equation in the interval $0\le x\le \frac{\pi }{2}$ .

Error:

In the first step both sides of the equation is divided by $\sin x$ , which is not correct as for some values of x , $\sin x=0$ and division by 0 is not defined.

Correct solution:

  $x=0$ and $x=\frac{\pi }{6}$

Given information:

The given solution is:

  $\begin{array}{c}{\sin }^{2}x=\frac{1}{2}\sin x\\ \sin x=\frac{1}{2}\\ x=\frac{\pi }{6}\end{array}$

Explanation:

The mistake in the given solution is that in the first step both sides of the equation is divided by $\sin x$ , which is not correct as for some values of x , $\sin x=0$ and division by 0 is not defined.

The correct solution is:

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

Now, set each factor equal to 0, and solve for x .

  $\begin{array}{c}\sin x=0\\ x=0,0\le x\le \frac{\pi }{2}\end{array}$

Equating the other factor to 0 gives,

  $\begin{array}{c}\sin x-\frac{1}{2}=0\\ \sin x=\frac{1}{2}\\ x=\frac{\pi }{6},0\le x\le \frac{\pi }{2}\end{array}$

Therefore the solution is $x=0$ and $x=\frac{\pi }{6}$ .