I study maths as a hobby. I have this problem:
Find in radians the general solution of:
$$\cos x + \cos 3x + \cos 5x = 0$$
I have said:
$\cos 3x + \cos 5x = 2\cos\left( \frac{1}{2}(3x + 5x)\right)\cos\left( \frac{1}{2} (3x – 5x)\right) = 2\cos 4x\cos x$
$\cos x + 2\cos 4x \cdot \cos (x) = 0$
$(1 + 2\cos 4x)\cos x = 0$
$\cos x = 0$ or $\cos 4x = – \frac{1}{2}$
$x = \frac{\pi}{2}$ or $x = \frac{\pi}{3} $
My book gives the answers as:
$(2n + 1)\frac{\pi}{6}, n\pi \pm \frac{\pi}{3}$
So we agree on the second answer but is the book wrong when it says $(2n + 1)\frac{\pi}{6}$?
I get the answer as $x = n\pi + \frac{\pi}{2}$, which can be written as $(2n +1)\frac{\pi}{2}$
Best Answer
$$\cos 4x =-1/2 \iff \cos4x +\cos \frac{\pi}{3} =0 \iff 2\cos\left(2x +\frac{\pi}{6} \right)\cos\left(2x-\frac{\pi}{6} \right)=0 \\ \iff \cos\left(2x +\frac{\pi}{6} \right) = 0 \text{ or } \cos\left(2x -\frac{\pi}{6} \right) = 0 \\ \iff 2x +\frac{\pi}{6} = \frac{\pi}{2}+n\pi \text{ or } 2x -\frac{\pi}{6} = \frac{\pi}{2}+n\pi \\ \iff 2x = \frac{\pi}{3}+n\pi \text{ or } 2x = \frac{2\pi}{3}+n\pi\\ \iff x=\frac{\pi}{6} + n\frac{\pi}{2} \text{ or }x= \frac{\pi}{3}+n\frac{\pi}{2} \\ \iff x= (3n+1) \frac{\pi}{6} \text{ or } x= (3n+2) \frac{\pi}{6}$$ And as you said, for $\cos x=0$ you need to take $$x=(2n+1) \frac{\pi}{2} = (6n+3)\frac{\pi}{6}$$
This is, in fact, equivalent to the book’s answer. To see this, observe that the union of $3n+1, 3n+2, ,6n+3$ is all the integers $m$ such that $m\equiv 1,2,3,4,5 \pmod 6$, which is further equivalent to the union of the odd numbers and numbers of the form $6n+2$ and $6n-2$. So, the solution set can be written as $$x=(2n+1)\frac{\pi}{6} \ \text{or} \ (6n\pm 2)\frac{\pi}{6} = (2n+1)\frac{\pi}{6} \ \text{or} \ n\pi \pm \frac{\pi}{3}$$