Question

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