Chapter 7.5, Problem 19E

a.

To determine

To solve : the given equation $\{\text{ 2}\cos 2\theta +1=0$ .

Answer to Problem 19E

The general solutions of $\theta$ for the equation $\{\text{ 2}\cos 2\theta +1=0$ are

  $\theta =\frac{\pi }{3}\text{+2n}\pi ,\theta \text{=}\frac{2\pi }{3}+2n\pi ,\theta =\frac{4\pi }{3}\text{+2n}\pi \text{ or }\theta \text{=}\frac{5\pi }{3}+2n\pi$ .

Explanation of Solution

Given information:

The equation to be solved is $\{\text{ 2}\cos 2\theta +1=0$ .

Calculation :

  $\text{2}\cos 2\theta +1=0$

Solve the given equation.

  $\begin{array}{l}2\left(1-2{\sin }^2\theta \right)+1=0\\ 2-4{\sin }^2\theta +1=0\\ 3=4{\sin }^2\theta \\ \sin \theta =\pm \frac{\sqrt{3}}{2}\end{array}$

Now, there are two cases. Solve each case to get the solution of $\theta$ .

Case 1:

  $\sin \theta =\frac{\sqrt{3}}{2}$

Above equation is satisfied in the interval $\left[0,2\pi \right]$ when $\theta =\frac{\pi }{3}\text{ or }\theta \text{=}\frac{2\pi }{3}$ .

Since, the period of sine function is $2\pi$ . Add $2\pi n$ to get the general solution of $\theta$ where $n$ is any integer.

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

Case 2:

  $\sin \theta =\frac{-\sqrt{3}}{2}$

Above equation is satisfied in the interval $\left[0,2\pi \right]$ when $\theta =\frac{4\pi }{3}\text{ or }\theta \text{=}\frac{5\pi }{3}$ .

Since, the period of sine function is $2\pi$ . Add $2\pi n$ to get the general solution of $\theta$ where $n$ is any integer.

  $\theta =\frac{4\pi }{3}\text{+2n}\pi \text{ or }\theta \text{=}\frac{5\pi }{3}+2n\pi$

Therefore, the general solutions of $\theta$ for the equation $\{\text{ 2}\cos 2\theta +1=0$ are

  $\theta =\frac{\pi }{3}\text{+2n}\pi ,\theta \text{=}\frac{2\pi }{3}+2n\pi ,\theta =\frac{4\pi }{3}\text{+2n}\pi \text{ or }\theta \text{=}\frac{5\pi }{3}+2n\pi$ .

b.

To determine

To find: the solutions of the equation $\{\text{ 2}\cos 2\theta +1=0$ in the interval $[0,2\pi )$ .

Answer to Problem 19E

The general solutions of equation are $\{\text{ 2}\cos 2\theta +1=0$ in the interval $[0,2\pi )$ are

  $\theta =\frac{\pi }{3}\text{, }\theta =\frac{2\pi }{3},\theta =\frac{4\pi }{3}\text{ and }\theta =\frac{5\pi }{3}$

Explanation of Solution

Given information:

The equation to be solved is $\{\text{ 2}\cos 2\theta +1=0$ in the interval $[0,2\pi )$ .

Calculation:

The general solutions of $\theta$ for the equation $\{\text{ 2}\cos 2\theta +1=0$ are

  $\theta =\frac{\pi }{3}\text{+2n}\pi ,\theta \text{=}\frac{2\pi }{3}+2n\pi ,\theta =\frac{4\pi }{3}\text{+2n}\pi \text{ or }\theta \text{=}\frac{5\pi }{3}+2n\pi$ .

Put $n=0,1,2,\mathrm{3......}$ in such a way that $\theta$ should not exceed $2\pi$ because $\theta \in [0,2\pi )$ .

At $n=0$ :

  $\theta =\frac{\pi }{3}\text{, }\theta =\frac{2\pi }{3},\theta =\frac{4\pi }{3}\text{ and }\theta =\frac{5\pi }{3}$

At $n=1$ :

  $\theta =\frac{7\pi }{3}\text{, }\theta =\frac{8\pi }{3},\theta =\frac{10\pi }{3}\text{ and }\theta =\frac{11\pi }{3}$

Since, solutions at $n=1$ will not be considered because $\theta >2\pi$ .

Therefore, the general solutions of equation are $\{\text{ 2}\cos 2\theta +1=0$ in the interval $[0,2\pi )$ are

  $\theta =\frac{\pi }{3}\text{, }\theta =\frac{2\pi }{3},\theta =\frac{4\pi }{3}\text{ and }\theta =\frac{5\pi }{3}$ .