Chapter 7.5, Problem 44E

To determine

To find: The solution of equation $\tan \frac{\theta }{2}-\sin \theta =0$ in the interval $\left[0,2\pi \right]$ .

Answer to Problem 44E

The solution of the equation $\tan \frac{\theta }{2}-\sin \theta =0$ in interval are $0,\frac{\pi }{2}$ and $\frac{3\pi }{2}$ .

Explanation of Solution

Given information:

The equation is $\tan \frac{\theta }{2}-\sin \theta =0$ .

Calculation:

Write the double angle formula.

  $\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$

Simplify the equation using double angle formula

  $\begin{array}{c}\tan \frac{\theta }{2}-\sin \theta =0\\ \tan \frac{\theta }{2}-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=0\\ \frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=0\end{array}$..............(1)

Multiply both sides by $\cos \frac{\theta }{2}$ .

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

Further solve.

  $\begin{array}{l}\sin \frac{\theta }{2}=0\\ \theta =2n\pi \end{array}$

Solve for $1-2{\cos }^2\frac{\theta }{2}=0$ .

  $\begin{array}{c}1-2{\cos }^2\frac{\theta }{2}=0\\ {\cos }^2\frac{\theta }{2}=\frac{1}{2}\\ \cos \frac{\theta }{2}=\pm \frac{1}{\sqrt{2}}\end{array}$

Solve for positive value.

  $\theta =\frac{\pi }{2}\text{or}\frac{7\pi }{2}$

Solve for negative value.

  $\theta =\frac{3\pi }{2}\text{or}\frac{5\pi }{2}$

  $\theta =\frac{\pi }{2}\text{or}\frac{7\pi }{2}$

Since the value is to be calculated in the range of $\left[0,2\pi \right]$ . So, $\frac{5\pi }{2}$ and $\frac{7\pi }{2}$ will not be considered.

Therefore, the solution of the equation $\tan \frac{\theta }{2}-\sin \theta =0$ in interval are $0,\frac{\pi }{2}$ and $\frac{3\pi }{2}$ .