Chapter 7.4, Problem 40E

To determine

To find: The solution of the equation $\left(\tan \theta -2\right)\left(16{\sin }^2\theta -1\right)=0$ .

Answer to Problem 40E

The solutions of the equation $\left(\tan \theta -2\right)\left(16{\sin }^2\theta -1\right)=0$ are $\theta =-1.1+k\pi$ , $\theta =1.1+k\pi$ , $\theta =\frac{2\pi }{3}+2k\pi$ and $\theta =\frac{4\pi }{3}+2k\pi$ .

Explanation of Solution

Given information:

The given equation is $\left(\tan \theta -2\right)\left(16{\sin }^2\theta -1\right)=0$ .

Calculation:

First substitute the factor $\left(\tan \theta -2\right)=0$ .

Simplify the equation

  $\begin{array}{l}\tan \theta -2=0\\ \tan \theta =2\end{array}$

As tangent has period $\pi$ . So first find the solutions in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ .

Therefore, the solution to the equation in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ are

  $\begin{array}{c}\theta ={\tan }^{-1}\left(2\right)\\ \approx 1.1\end{array}$

As tangent repeats its values after every $\pi$ units.

So, the general solution of the equation is given by

  $\theta =1.1+k\pi$

Therefore, the solutions of the equation $\tan \theta -2=0$ are $\theta =1.1+k\pi$ .

Now substitute the factor $16{\sin }^2\theta -1=0$ .

Simplify the equation

  $\begin{array}{c}16{\sin }^2\theta -1=0\\ 16{\sin }^2\theta =1\\ {\sin }^2\theta =\frac{1}{16}\\ \sin \theta =\pm \frac{1}{4}\end{array}$

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

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1}{4}\right)\\ \approx 0.25,2.89\end{array}$

And,

  $\begin{array}{c}\theta ={\sin }^{-1}\left(-\frac{1}{4}\right)\\ \approx 3.39,6.03\end{array}$

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

So, the general solution of the equation is given by

  $\theta =0.25+2k\pi ,2.89+2k\pi ,3.39+2k\pi ,6.03+2k\pi$

Therefore, the solutions of the equation $16{\sin }^2\theta -1=0$ are $\theta =0.25+k\pi$ and $\theta =3.39+k\pi$ .

Combining the solution of both the factors gives the solution of the equation $\left(\tan \theta -2\right)\left(16{\sin }^2\theta -1\right)=0$ .

Therefore, the solutions of the equation $\left(\tan \theta -2\right)\left(16{\sin }^2\theta -1\right)=0$ are $\theta =1.1+k\pi$ , $\theta =0.25+k\pi$ and $\theta =3.39+k\pi$ .