Chapter 10.4, Problem 24E

Solution

To find: the general solution of the given equation.

  $x=\frac{\pi }{2}+2n\pi$ or $x=\frac{3\pi }{2}+2n\pi$ , where $n$ is any integer.

Given:

  $\sin x\cos x-3\cos x=0$

Concept used:

The sine function is positive in I and II quadrants.

The cosine function is positive in I and IV quadrants.

The sine function is periodic with a period of $2\pi$ .

The cosine function is periodic with a period of $2\pi$ .

Calculation:

Consider,

  $\begin{array}{l}\sin x\cos x-3\cos x=0\\ \cos x\left(\sin x-3\right)=0\\ \cos x=0\text{or}\sin x-3=0\\ \cos x=0\text{or}\sin x=3\end{array}$

But $\sin x=3$ is not possible, because the range of sine function is $\left[-1,1\right]$ .

Therefore,

  $\cos x=0$

Now, one solution of $\cos x=0$ in the interval $0\le x<2\pi$ is,

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

The other solution in the interval is,

  $\begin{array}{c}x=2\pi -\frac{\pi }{2}\\ =\frac{3\pi }{2}\end{array}$

Now since the cosine function is periodic with a period of $2\pi$ , therefore the general solution is,

  $x=\frac{\pi }{2}+2n\pi$ or $x=\frac{3\pi }{2}+2n\pi$ , where $n$ is any integer.