Question

If $0 \leq x < 2 \pi$ , then the number of real values of $x$, which satisfy the equation
$\cos x + \cos 2x + \cos 3x + \cos 4x = 0$, is :

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

We have, $\cos x + \cos 2x + \cos 3x+\cos 4x = 0$
$(\cos x + \cos 4 x) + (\cos 2x + \cos 3x ) = 0$
$2 \cos \frac{5x}{2} . \cos \frac{3x}{2} + 2 \cos \frac{5x}{2} . \cos \frac{x}{2} = 0 $
$ 2 \cos \frac{5x}{2} \left[\cos \frac{3x}{2} + \cos \frac{x}{2}\right] = 0$
$ \cos \frac{5x}{2} = 0 \Rightarrow \frac{5x}{2} =\left(2n+1\right) \frac{\pi}{2} \Rightarrow x = \left(2n+1\right) \frac{\pi}{5} \Rightarrow x = \frac{\pi}{5}, \frac{3\pi}{5} , \pi, \frac{7 \pi}{5} , \frac{9\pi}{5} $
$\cos \frac{3x}{2} + \cos \frac{x}{2} = 0$
$ 4 \cos^{3} \frac{x}{2} - 3 \cos \frac{x}{2} + \cos \frac{x}{2} = 0 \Rightarrow 4 \cos^{3} -2 \cos \frac{x}{2} = 0 \Rightarrow 2 \cos^{3} \frac{x}{2} - \cos \frac{x}{2} = 0$
$ 2 \cos \frac{x}{2} \left[2 \cos^{2} \frac{x}{2} - 1\right] = 0 \Rightarrow 2 \cos \frac{x}{2} \left[\cos x \right] = 0$
$ \cos \frac{x}{2} = 0 \Rightarrow \frac{x}{2} = \left(2n + 1\right) \frac{\pi}{2} \Rightarrow x =\left(2n+1\right)\pi $ or $\cos x = 0 \Rightarrow x = \left(2n+1\right) \frac{\pi}{2}$
$ x = \pi$ or $x = \frac{\pi}{2}, \frac{3\pi}{2} $
Solution are $x = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5} , \frac{9 \pi}{5}, \frac{\pi}{2} , \frac{3\pi}{2} .....\left(0 \le x < 2 \pi\right) $