Chapter 7, Problem 7T

(a)

To determine

To find: The solution of the equation $3\sin \theta -1=0$ within the interval $\left[0,2\pi \right)$ .

Answer to Problem 7T

The solution of the equation $3\sin \theta -1=0$ within the interval $\left[0,2\pi \right)$ is $\theta =0\cdot 34,2\cdot 80$ .

Explanation of Solution

Given information:

The given trigonometric equation is $3\sin \theta -1=0$ .

Calculation:

Isolate the term $\sin \theta$ and apply the inverse function to obtain the values of $\theta$ . The graph with unit circle and the line $\frac{1}{3}$ is shown in figure (1).

  

Figure (1)

Calculate the value of $\sin \theta$ .

  $\begin{array}{c}3\sin \theta -1=0\\ \sin \theta =\frac{1}{3}\end{array}$

Using arcsine on both sides.

  $\begin{array}{l}\theta ={\sin }^{-1}\left(\frac{1}{3}\right)\\ \theta =0\cdot 34,2\cdot 80\end{array}$

Therefore, the solution of the equation $3\sin \theta -1=0$ within the interval are $\left[0,2\pi \right)$ is $\theta =0\cdot 34,2\cdot 80$ .

(b)

To determine

To find: The solution of the equation $\left(2\cos \theta -1\right)\left(\sin \theta -1\right)=0$ within the interval $\left[0,2\pi \right)$ .

Answer to Problem 7T

The solution of the equation $\left(2\cos \theta -1\right)\left(\sin \theta -1\right)=0$ within the interval $\left[0,2\pi \right)$ are $\theta =\frac{\pi }{3},-\frac{\pi }{3}$ and $\frac{\pi }{2}$ .

Explanation of Solution

Given information:

The given trigonometric equation is $\left(2\cos \theta -1\right)\left(\sin \theta -1\right)=0$ .

Calculation:

For $2\cos \theta -1=0$ , isolate the term $\cos \theta$ and apply the inverse cosine function.

Solve the equation.

  $\begin{array}{c}2\cos \theta -1=0\\ \cos \theta =\frac{1}{2}\\ \theta ={\cos }^{-1}\left(\frac{1}{2}\right)\\ \theta =\frac{\pi }{3},-\frac{\pi }{3}\end{array}$

For $\sin \theta -1=0$ , isolate the term $\sin \theta$ and apply the inverse sine function.

Solve the equation.

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

Therefore, the solution of the equation $\left(2\cos \theta -1\right)\left(\sin \theta -1\right)=0$ within the interval $\left[0,2\pi \right)$ are $\theta =\frac{\pi }{3},-\frac{\pi }{3}$ and $\frac{\pi }{2}$ .

(c)

To determine

To find: The solution of the equation $2{\cos }^2\theta +5\cos \theta +2=0$ within the interval $\left[0,2\pi \right)$ .

Answer to Problem 7T

The solution of the equation $2{\cos }^2\theta +5\cos \theta +2=0$ within the interval $\left[0,2\pi \right)$ is $\theta =\frac{2\pi }{3},\frac{4\pi }{3}$ .

Explanation of Solution

Given information:

The given trigonometric equation is $2{\cos }^2\theta +5\cos \theta +2=0$ .

Calculation:

Solve the value of $\theta$ .

  $\begin{array}{c}2{\cos }^2\theta +5\cos \theta +2=0\\ \left(2\cos \theta +1\right)\left(\cos \theta +2\right)=2\end{array}$

Using the following property.

If $x\cdot y=0$ , then $x=0,y=0$ .

Solve for $2\cos \theta +1=0$ .

  $\begin{array}{c}2\cos \theta +1=0\\ \cos \theta =-\frac{1}{2}\\ \theta ={\cos }^{-1}\left(-\frac{1}{2}\right)\\ \theta =\frac{2\pi }{3},\frac{4\pi }{3}\end{array}$

The angles are $\theta =\frac{2\pi }{3},\frac{4\pi }{3}$ since cosine is negative in quadrant $\text{I}$ and quadrant $\text{III}$ .

If $\cos \theta +2=0$ .

  $\begin{array}{c}\cos \theta +2=0\\ \cos \theta =-2\end{array}$

As the cosine function range from $-1$ to $1$ , it does not give us a solution. The solution for this equation lies outside the range so these are excluded.

Therefore, the solution of the equation $2{\cos }^2\theta +5\cos \theta +2=0$ within the interval $\left[0,2\pi \right)$ is $\theta =\frac{2\pi }{3},\frac{4\pi }{3}$ .

(d)

To determine

To find: The solution of the equation $\sin 2\theta -\cos \theta =0$ within the interval $\left[0,2\pi \right)$ .

Answer to Problem 7T

The solution of the equation $\sin 2\theta -\cos \theta =0$ within the interval $\left[0,2\pi \right)$ is $\theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}$ .

Explanation of Solution

Given information:

The given trigonometric equation is $\sin 2\theta -\cos \theta =0$ .

Calculation:

Using the double angle formula for sine.

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

Sine takes positive values in quadrant $\text{I}$ and $\text{II}$ , the two possible solutions are as follows.

  $\theta =\frac{\pi }{6},\frac{5\pi }{6}$

The factor equal to zero.

  $\begin{array}{c}\cos \theta =0\\ \theta =\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

Therefore, the solution of the equation $\sin 2\theta -\cos \theta =0$ within the interval $\left[0,2\pi \right)$ is $\theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}$ .