Chapter 5.3, Problem 36E

Solution

To solve: The equation $\cos 3x\cos x=\sin 3x\sin x$ by using sum or difference identities.

The solution of the given equation is $x=\frac{\pi }{8}+\frac{n\pi }{2}$ .

Given information:

The equation is $\cos 3x\cos x=\sin 3x\sin x$ .

Formula used:

Use the formula for cosine of sum.

  $\cos \left(u+v\right)=\cos u\cos v-\sin u\sin v$

Calculation:

Subtract the $\sin 3x\sin x$ from the R.H.S. and rewrite the equation in standard form.

  $\cos 3x\cos x-\sin 3x\sin x=0$

Use the cosine of a sum formula.

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

Let $u=4x$ . On $\left[0,2\pi \right)$ , cosine is $0$ for angles $\frac{\pi }{2}$ and $\frac{3\pi }{2}$ . The period of $\cos u$ is $2\pi$ . So,

  $\begin{array}{c}u=\frac{\pi }{2}+2n\pi \text{or}\frac{3\pi }{2}+2n\pi \\ 4x=\frac{\pi }{2}+2n\pi \text{or}\frac{3\pi }{2}+2n\pi \\ x=\frac{\pi }{8}+\frac{n\pi }{2}\text{or}\frac{3\pi }{8}+\frac{n\pi }{2}\end{array}$

Here, $n$ is an integer which is $n=0,\pm 1,\pm 2,\mathrm{...}$ .

Combine the two solutions into one since they are composed of multiples of $\frac{\pi }{8}$ , replace the period by $\frac{\pi }{4}$ .

  $x=\frac{\pi }{8}+\frac{n\pi }{2}$

Therefore, the solution of the given equation is $x=\frac{\pi }{8}+\frac{n\pi }{2}$ .