Chapter 5, Problem 61CR

To determine

To calculate: The solutions of the multiple-angle equation.

Answer to Problem 61CR

The required solutions are $x=\frac{\pi }{12}+\frac{n\pi }{6}$ , for $n=0,1,2,\cdots$ .

Explanation of Solution

Given information:

The equation: $2{\sin }^{2}3x-1=0$

Calculation:

The given equation is $2{\sin }^{2}3x-1=0$ .

Rewrite the above equation as:

  $\begin{array}{c}2{\sin }^{2}3x-1=0\\ \frac{\cancel{2}{\sin }^{2}3x}{\cancel{2}}=\frac{1}{2}\\ \sqrt{{\sin }^{2}3x}=\pm \frac{1}{\sqrt{2}}\\ \sin 3x=\pm \frac{1}{\sqrt{2}}\end{array}$

This means $\sin 3x=\pm \frac{1}{\sqrt{2}}$ at,

  $\begin{array}{c}\frac{3x}{3}=\frac{\pi }{4\times 3}+\frac{2n\pi }{3}\\ x=\frac{\pi }{12}+\frac{2n\pi }{3}\end{array}$ ,

  $\begin{array}{c}\frac{3x}{3}=\frac{3\pi }{4\times 3}+\frac{2n\pi }{3}\\ x=\frac{3\pi }{12}+\frac{2n\pi }{3}\end{array}$ ,

And, $\begin{array}{c}\frac{3x}{3}=\frac{5\pi }{4\times 3}+\frac{2n\pi }{3}\\ x=\frac{5\pi }{12}+\frac{2n\pi }{3}\end{array}$

And so on.

Thus, the equation has all solution $x=\frac{\pi }{12}+\frac{n\pi }{6}$ , for $n=0,1,2,\cdots$ .

Hence, the required solutions are $x=\frac{\pi }{12}+\frac{n\pi }{6}$ , for $n=0,1,2,\cdots$ .