Chapter 7.5, Problem 32E

(a)

To determine

To find: The solution of the equation $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ .

Answer to Problem 32E

The solutions of the equation $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ are $\theta =\frac{\pi }{6}+2k\pi$ , $\theta =\frac{5\pi }{6}+2k\pi$

  $\theta =\frac{2\pi }{3}+2k\pi$ and $\theta =\frac{4\pi }{3}+2k\pi$ .

Explanation of Solution

Given information:

The given equation is $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ .

Calculation:

Simplify the equation.

  $\begin{array}{c}4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0\\ 2\sin \theta \left(2\cos \theta +1\right)-1\left(2\cos \theta +1\right)=0\\ \left(2\sin \theta -1\right)\left(2\cos \theta +1\right)=0\end{array}$

First substitute the factor $2\sin \theta -1=0$ .

Simplify the equation.

  $\begin{array}{c}2\sin \theta -1=0\\ 2\sin \theta =1\\ \sin \theta =\frac{1}{2}\end{array}$

As sine 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

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1}{2}\right)\\ =\frac{\pi }{6},\frac{5\pi }{6}\end{array}$

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

So, the general solution of the equation is given by

  $\theta =\frac{\pi }{6}+2k\pi ,\frac{5\pi }{6}+2k\pi$

Now substitute the factor

  $2\cos \theta +1=0$

Simplify the equation.

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

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

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

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

So, the general solution of the equation is given by

  $\theta =\frac{2\pi }{3}+2k\pi ,\frac{4\pi }{3}+2k\pi$

Therefore, the solutions of the equation $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ are $\theta =\frac{\pi }{6}+2k\pi$ , $\theta =\frac{5\pi }{6}+2k\pi$

  $\theta =\frac{2\pi }{3}+2k\pi$ and $\theta =\frac{4\pi }{3}+2k\pi$ .

(b)

To determine

To find: The solution of the equation $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ in $\left[0,2\pi \right)$ .

Answer to Problem 32E

The solution of the equation $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ in $\left[0,2\pi \right)$ are $\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{2\pi }{3},\frac{4\pi }{3}$ .

Explanation of Solution

Given information:

The given equation is $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ and the interval is $\left[0,2\pi \right)$ .

Calculation:

According to solution in part(a) the general solution of the equation $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ are $\theta =\frac{\pi }{6}+2k\pi$ , $\theta =\frac{5\pi }{6}+2k\pi$$\theta =\frac{2\pi }{3}+2k\pi$ and $\theta =\frac{4\pi }{3}+2k\pi$ . The solution of the equation in $\left[0,2\pi \right)$ corresponds to $k=0$ i.e. $\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{2\pi }{3},\frac{4\pi }{3}$ .

Therefore, the solutions of the equation $4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1=0$ in $\left[0,2\pi \right)$ are $\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{2\pi }{3},\frac{4\pi }{3}$ .