Question

If $0 \leqslant x < 2\pi$ , then the number of real values of x , which satisfy the equation $\cos x + \cos 2x + \cos 3x + \cos 4x = 0$;
A. $3$
B. $5$
C. $7$
D. $9$

Answer 1

In the above question, the given expression is $\cos x + \cos 2x + \cos 3x + \cos 4x = 0$ , we need to find the value of x which satisfies this expression. So, solve the expression using $\cos a + \cos b = 2\cos \dfrac{{(a + b)}}{2}\cos \dfrac{{(a - b)}}{2}$ trigonometric formula and get the required answer.

Complete step-by-step answer:
According to the question we need to find numbers of x that will satisfy both the equations.
Consider (1)
Given $0 \leqslant x < 2\pi .......................(1)$
$\cos x + \cos 2x + \cos 3x + \cos 4x = 0......................(2)$
$\cos x + \cos 2x + \cos 3x + \cos 4x = 0 \\\ \cos x + \cos 3x + \cos 2x + \cos 4x = 0 \\\ {\text{Now we can say that}} \\\ \cos a + \cos b = 2\cos \dfrac{{a + b}}{2}\cos \dfrac{{a - b}}{2}....(3) \\\ 2\cos \dfrac{{(x + 3x)}}{2}\cos \dfrac{{(x - 3x)}}{2} + 2\cos \dfrac{{(2x + 4x)}}{2}\cos \dfrac{{(2x - 4x)}}{2} = 0 \\\ 2\cos 2x\cos ( - x) + 2\cos (3x)\cos ( - x) = 0 \\\ {\text{Now, }}\cos ( - a) = \cos (a) \\\$
$4\cos x\cos \dfrac{{5x}}{2}\cos \dfrac{{ - x}}{2} = 0$
As we know that $\cos ( - a) = \cos a$, so we can write
$2\cos 2x\cos ( x) + 2\cos (3x)\cos ( x) = 0$
Taking $2\cos x$ common we get
$2\cos x (\cos 2x + \cos 3x)$
Again applying the formula $\cos a + \cos b = 2\cos \dfrac{{a + b}}{2}\cos \dfrac{{a - b}}{2}$ we get,
$2\cos x (2\cos \dfrac{{2x + 3x }}{2}\cos \dfrac{{2x - 3x}}{2})$
$2\cos x (2\cos \dfrac{{5x }}{2}\cos \dfrac{{ - x}}{2})$
$4\cos x\cos \dfrac{{5x}}{2}\cos \dfrac{x}{2} = 0$
$\cos x = 0;\cos \dfrac{x}{2} = 0;\cos \dfrac{{5x}}{2} = 0 \\\ {\text{Use ,}}\cos (2n + 1)\dfrac{\pi }{2} = 0;n \in \mathbb{Z} \\\ \cos x = \cos (2n + 1)\dfrac{\pi }{2};\cos \dfrac{x}{2} = \cos (2m + 1)\dfrac{\pi }{2};\cos \dfrac{{5x}}{2} = \cos (2k + 1)\dfrac{\pi }{2};n,k,m \in \mathbb{Z} \\\ x = (2n + 1)\dfrac{\pi }{2};\dfrac{x}{2} = (2m + 1)\dfrac{\pi }{2};\dfrac{{5x}}{2} = (2k + 1)\dfrac{\pi }{2} \\\$
Now, we know that $0 \leqslant x < 2\pi$
Now taking
$x = (2n + 1)\dfrac{\pi }{2},n \in \mathbb{Z} \\\ {\text{When; }}n = 0 = > x = \dfrac{\pi }{2} \\\ n = 1 = > x = \dfrac{{3\pi }}{2} \\\ So,x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}.........................(4) \\\$
Now taking
$\dfrac{x}{2} = (2m + 1)\dfrac{\pi }{2},k \in \mathbb{Z} \\\ x = (2m + 1)\pi \\\ {\text{When, }}m = 0 = > x = \pi ..............(5) \\\$
Now taking ,
$\dfrac{{5x}}{2} = (2k + 1)\dfrac{\pi }{2},k \in \mathbb{Z} \\\ x = (2k + 1)\dfrac{\pi }{5} \\\ {\text{When,}} \\\ k = 0, = > x = \dfrac{\pi }{5} \\\ k = 1, = > x = \dfrac{{3\pi }}{5} \\\ k = 2, = > x = \pi \\\ k = 3, = > x = \dfrac{{7\pi }}{5} \\\ k = 4, = > x = \dfrac{{9\pi }}{5} \\\ so, \\\ x = \dfrac{\pi }{5},\dfrac{{3\pi }}{5},\pi ,\dfrac{{7\pi }}{5},\dfrac{{9\pi }}{5}.........................(6) \\\$
Now , from (4), (5), (6)
$x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{\pi }{5},\dfrac{{3\pi }}{5},\pi ,\dfrac{{7\pi }}{5},\dfrac{{9\pi }}{5}$………………..
So, there are seven values of x satisfying (1) and (2)

So, the correct answer is “Option C”.

Note: Alternative method to solve their question is
$\cos x + \cos 2x + \cos 3x + \cos 4x = 0$
We can also use (3) for $\cos x + \cos 2x;\cos 3x + \cos 4x,$like this,
$2\cos \dfrac{{3x}}{2}\cos \dfrac{x}{2} + 2\cos \dfrac{{7x}}{2}\cos \dfrac{x}{2} = 0 \\\ 2\cos \dfrac{x}{2}(\cos \dfrac{{3x}}{2} + \cos \dfrac{{7x}}{2}) = 0 \\\ \\\$
Now, again using (3)
$2\cos \dfrac{x}{2}(2\cos 5x\cos 2x) = 0 \\\ cox\dfrac{x}{2} = 0;\cos 2x = 0;\cos 5x = 0 \\\$
Similarly find the real values of x.