Chapter 7.4, Problem 56E

To determine

To find: The solution of the equation $4\cos \theta \sin \theta +3\cos \theta =0$ .

Answer to Problem 56E

The solutions of the equation $4\cos \theta \sin \theta +3\cos \theta =0$ are $\theta =\frac{\pi }{2}+k\pi$ and $\theta =3.99+2k\pi$ and $\theta =5.44+2k\pi$ .

Explanation of Solution

Given information:

The given equation is $4\cos \theta \sin \theta +3\cos \theta =0$ .

Calculation:

Simplify the equation.

  $\begin{array}{c}4\cos \theta \sin \theta +3\cos \theta =0\\ \cos \theta \left(4\sin \theta +3\right)=0\end{array}$

First substitute the factor $\cos \theta =0$ .

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

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

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

So, the general solution of the equation is given by

  $\begin{array}{c}\theta =\frac{\pi }{2}+2k\pi ,\frac{3\pi }{2}+2k\pi \\ =\frac{\pi }{2}+k\pi \end{array}$

Therefore, the solutions of the equation $\cos \theta =0$ are $\theta =\frac{\pi }{2}+k\pi$ .

Now substitute the factor $4\sin \theta +3=0$ .

Simplify the equation.

  $\begin{array}{c}4\sin \theta +3=0\\ 4\sin \theta =-3\\ \sin \theta =-\frac{3}{4}\end{array}$

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

  $\begin{array}{c}\theta ={\sin }^{-1}\left(-\frac{3}{4}\right)\\ \approx 3.99,5.44\end{array}$

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

So, the general solution of the equation is given by

  $\theta =3.99+2k\pi ,5.44+2k\pi$

Therefore, the solutions of the equation $4\sin \theta +3=0$ are $\theta =3.99+2k\pi$ and $\theta =5.44+2k\pi$ .

Therefore, the solutions of the equation $4\cos \theta \sin \theta +3\cos \theta =0$ are $\theta =\frac{\pi }{2}+k\pi$ and $\theta =3.99+2k\pi$ and $\theta =5.44+2k\pi$ .