Chapter 10, Problem 18T

Solution

To calculate: The general solution of the equation $\sin \frac{x}{2}=1-\cos x$.

The resultant answer is $x=\frac{\pi }{3}+4n\pi \text{or}\frac{5\pi }{3}+4n\pi ,x=4n\pi \text{or}2\pi +4n\pi$.

Given information:

The given equation is $\sin \frac{x}{2}=1-\cos x$.

Formula Used: Use the double angle formula of cosine function $\cos 2a=1-2{\sin }^{2}a$.

Calculation:

Consider the given equation: $\sin \frac{x}{2}=1-\cos x$

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

Use the double angle formula of sine function $\sin 2a=2\sin a\cos a$ ,

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

Take common $\sin \frac{x}{2}$ ,

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

Solve $1-2\sin \frac{x}{2}=0$ ,

  $\begin{array}{c}1-2\sin \frac{x}{2}=0\\ \sin \frac{x}{2}=\frac{1}{2}\\ \frac{x}{2}=\frac{\pi }{6}+2n\pi \text{or}\frac{5\pi }{6}+2n\pi \\ x=\frac{\pi }{3}+4n\pi \text{or}\frac{5\pi }{3}+4n\pi \end{array}$

Solve $\sin \frac{x}{2}=0$ ,

  $\begin{array}{c}\sin \frac{x}{2}=0\\ \frac{x}{2}=0+2n\pi \text{or}\pi +2n\pi \\ x=4n\pi \text{or}2\pi +4n\pi \end{array}$

Therefore, the general solution is $x=\frac{\pi }{3}+4n\pi \text{or}\frac{5\pi }{3}+4n\pi ,x=4n\pi \text{or}2\pi +4n\pi$.