Chapter 7.5, Problem 26E

(a)

To determine

To find: The solution of the equation $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ .

Answer to Problem 26E

The solutions of the equation $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ are $\theta =2k\pi$ .

Explanation of Solution

Given information:

The given equation is $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ .

Calculation:

Simplify the equation.

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

As cosine has period $2\pi$ . So first find the solutions in $\left[0,2\pi \right)$ .

Therefore, the solution to the equation in $\left[0,2\pi \right)$ are

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

And,

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

As cosine repeats its values after every $2\pi$ units.

So, the general solution of the equation is given by

  $\begin{array}{c}\frac{\theta }{2}=0+2k\pi ,\pi +2k\pi \\ \frac{\theta }{2}=k\pi \\ \theta =2k\pi \end{array}$

Therefore, the solutions of the equation $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ are $\theta =2k\pi$ .

(b)

To determine

To find: The solution of the equation $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ in $\left[0,2\pi \right)$ .

Answer to Problem 26E

The solution of the equation $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ in $\left[0,2\pi \right)$ is $\theta =0$ .

Explanation of Solution

Given information:

The given equation is $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ and the interval is $\left[0,2\pi \right)$ .

Calculation:

According to solution in part(a) the general solution of the equation $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ are $\theta =2k\pi$ . The solution of the equation in $\left[0,2\pi \right)$ corresponds to $k=0$ i.e. $\theta =0$ .

Therefore, the solution of the equation $\sec \frac{\theta }{2}=\cos \frac{\theta }{2}$ in $\left[0,2\pi \right)$ is $\theta =0$ .