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

The correct option is C $7$

$\text{Given equation is}\cos x+\cos 2x+\cos 3x+\cos 4x=0\Rightarrow (\cos x+\cos 3x)+(\cos 2x+\cos 4x)=0\Rightarrow 2\cos 2x\cos x+2\cos 3x\cos x=0\Rightarrow 2\cos x(\cos 2x+\cos 3x)=0\Rightarrow 2\cos x\left(2\cos \frac{5x}{2}\cos \frac{x}{2}\right)=0\Rightarrow \cos x\cdot \cos \frac{5x}{2}\cdot \cos \frac{x}{2}=0\Rightarrow \cos x=0\text{or}\cos \frac{5x}{2}=0\text{or}\cos \frac{x}{2}=0\text{Now,}\cos x=0\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2}[\because 0\le x<2\pi ]\cos \frac{5x}{2}=0\Rightarrow \frac{5x}{2}=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2},\frac{9\pi }{2},\frac{11\pi }{2},....\Rightarrow x=\frac{\pi }{5},\frac{3\pi }{5},\pi ,\frac{7\pi }{5},\frac{9\pi }{5}[\because 0\le x<2\pi ]\text{and}\cos \frac{x}{2}=0\Rightarrow \frac{x}{2}=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},......\Rightarrow x=\pi [\because 0\le x<2\pi ]\text{Hence,}x=\frac{\pi }{2},\frac{3\pi }{2},\pi ,\frac{\pi }{5},\frac{3\pi }{5},\frac{7\pi }{5},\frac{9\pi }{5}$