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Here I have another problem from our Telegram group.

We are talking about fastest way of determining which of the following options gives the general solution to $$\cos(x)+\cos(7x)+\cos(5x)+\cos(3x)=0$$

  1. $\dfrac{n\pi}{4}$
  2. $\dfrac{n\pi}{2}$
  3. $\dfrac{n\pi}{8}$
  4. none

Using the formula $$\cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$ into two pair, we got \begin{align} 2\cos(4x)\cos(3x)+2\cos(4x)cos(x)&=0 \tag1\\ \cos(4x)\cos(2x)\cos(x)&=0 \tag2\\ x&=\frac{n\pi}{8}. \tag3 \end{align}

Is there any other idea so that I can solve this question?

lIlIlIlIlIlIlIlIlI
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Ciup Titu
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3 Answers3

2

The 3 first choices contain $\pi$ as a solution, but $\pi$ is not a solution. The answer is 4.

Gribouillis
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    That's the truly fast solution (in fact immediate with $x=0$ or $x=2\pi$). –  Oct 24 '24 at 07:35
1

$$z+z^3+z^5+z^7+z^{-1}+z^{-3}+z^{-5}+z^{-7}=z\frac{z^8-1}{z^2-1}-z^{-1}\frac{z^{-8}-1}{z^{-2}-1}=\frac{z^9-z^{-7}}{z^2-1}.$$ so $$z=e^{i \pi/16}$$ makes a solution, but $$z=\pm1$$ not.

The answer is 4.

0

Recall that $\sum_{k=1}^{n}\cos((2k-1)x) = \frac{\sin(2nx)}{2\sin(x)}$ for $\sin(x) \neq 0$. So your equation becomes $\sin(8x)=0$ for $x \neq m\pi,\, m \in \mathbb{Z}$. And if $ x= m\pi$ then $\sum_{k=1}^{n}\cos((2k-1)x) =\sum_{k=1}^{n} (-1)^{m} \neq 0 $.

The solution set is thus $\{\frac{\pi m}{8}\in \mathbb{R}:m \in \mathbb{Z}, \, 8 \text{ doesn't divide } m \}$.

Robert Lee
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