The solution of frac{1}{2} + cos x + cos 2x + | vedclass

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Question

(A) Given equation: $\frac{1}{2} + \cos x + \cos 2x + \cos 3x + \cos 4x = 0$Multiply by $2 \sin(\frac{x}{2})$ on both sides:$\sin(\frac{x}{2}) + 2 \si

Vedclass · Accepted Answer

$x = \frac{2n\pi}{9}, n \in I, n 
eq 9m, m \in I$

Vedclass · Answer

(A) Given equation: $\frac{1}{2} + \cos x + \cos 2x + \cos 3x + \cos 4x = 0$Multiply by $2 \sin(\frac{x}{2})$ on both sides:$\sin(\frac{x}{2}) + 2 \sin(\frac{x}{2})(\cos x + \cos 2x + \cos 3x + \cos 4x) = 0$Using $2 \sin A \cos B = \sin(A+B) - \sin(B-A)$:$\sin(\frac{x}{2}) + (\sin(\frac{3x}{2}) - \sin(\frac{x}{2})) + (\sin(\frac{5x}{2}) - \sin(\frac{3x}{2})) + (\sin(\frac{7x}{2}) - \sin(\frac{5x}{2})) + (\sin(\frac{9x}{2}) - \sin(\frac{7x}{2})) = 0$This simplifies to $\sin(\frac{9x}{2}) = 0$,where $\sin(\frac{x}{2}) 
eq 0$.$\frac{9x}{2} = n\pi \Rightarrow x = \frac{2n\pi}{9}$.Since $\sin(\frac{x}{2}) 
eq 0$,$x 
eq 2k\pi$,which means $n$ cannot be a multiple of $9$ (i.e.,$n 
eq 9m$ for $m \in I$).

The solution of $\frac{1}{2} + \cos x + \cos 2x + \cos 3x + \cos 4x = 0$ is:

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