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

We have to count all real numbers $$x$$ lying in the interval $$0 \le x < 2\pi$$ for which the trigonometric equation

$$\cos x + \cos 2x + \cos 3x + \cos 4x = 0$$

is satisfied. To simplify the expression, we shall repeatedly use the well-known sum-to-product identity

$$\cos A + \cos B = 2\cos\!\left(\dfrac{A+B}{2}\right)\cos\!\left(\dfrac{A-B}{2}\right).$$

First we group the first and the last term, and also the two middle terms:

$$\bigl(\cos x + \cos 4x\bigr) + \bigl(\cos 2x + \cos 3x\bigr).$$

Applying the identity to each pair, we obtain

$$\cos x + \cos 4x = 2\cos\!\left(\dfrac{x+4x}{2}\right) \cos\!\left(\dfrac{x-4x}{2}\right) = 2\cos(2.5x)\cos(1.5x),$$

because $$\dfrac{x+4x}{2}=2.5x$$ and $$\dfrac{x-4x}{2}=-1.5x$$ while $$\cos(-\theta)=\cos\theta$$.

Similarly,

$$\cos 2x + \cos 3x = 2\cos\!\left(\dfrac{2x+3x}{2}\right) \cos\!\left(\dfrac{2x-3x}{2}\right) = 2\cos(2.5x)\cos(0.5x).$$

Adding the two results gives

$$\cos x + \cos 2x + \cos 3x + \cos 4x = 2\cos(2.5x)\cos(1.5x) + 2\cos(2.5x)\cos(0.5x).$$

Now we factor out the common factor $$2\cos(2.5x)$$:

$$\cos x + \cos 2x + \cos 3x + \cos 4x = 2\cos(2.5x)\Bigl[\cos(1.5x) + \cos(0.5x)\Bigr].$$

Inside the square brackets we again apply the same identity to the sum $$\cos(1.5x) + \cos(0.5x)$$:

$$\cos(1.5x) + \cos(0.5x) = 2\cos\!\left(\dfrac{1.5x+0.5x}{2}\right) \cos\!\left(\dfrac{1.5x-0.5x}{2}\right) = 2\cos(x)\cos(0.5x).$$

Substituting this back, we get

$$\cos x + \cos 2x + \cos 3x + \cos 4x = 2\cos(2.5x) \times 2\cos(x)\cos(0.5x) = 4\cos(2.5x)\cos(x)\cos(0.5x).$$

So the given equation becomes

$$4\cos(2.5x)\cos(x)\cos(0.5x) = 0.$$

The factor $$4$$ can never be zero, hence at least one of the three cosine factors must vanish. Therefore we have the three separate conditions

$$\text{(i)}\;\cos(2.5x)=0,\qquad \text{(ii)}\;\cos x=0,\qquad \text{(iii)}\;\cos(0.5x)=0.$$

We now solve each condition within $$0 \le x < 2\pi$$.

Condition (ii): $$\cos x = 0.$$

The cosine function is zero when its argument equals $$\dfrac{\pi}{2} + k\pi$$, where $$k$$ is an integer. Thus

$$x = \dfrac{\pi}{2} + k\pi.$$

Checking allowable values of $$k$$:

For $$k = 0$$, $$x = \dfrac{\pi}{2}.$$

For $$k = 1$$, $$x = \dfrac{\pi}{2} + \pi = \dfrac{3\pi}{2}.$$

For $$k = 2$$, $$x = \dfrac{\pi}{2} + 2\pi = \dfrac{5\pi}{2} > 2\pi,$$ so it is outside the desired interval.

Hence condition (ii) contributes the two solutions

$$x = \dfrac{\pi}{2},\; \dfrac{3\pi}{2}.$$

Condition (iii): $$\cos(0.5x) = 0.$$

Setting $$0.5x = \dfrac{\pi}{2} + k\pi$$ (again with integer $$k$$) we get

$$x = \pi + 2k\pi.$$

For $$k = 0$$, $$x = \pi$$ lies in the interval.

For $$k = 1$$, $$x = \pi + 2\pi = 3\pi > 2\pi,$$ so it is excluded.

Thus condition (iii) contributes the single solution

$$x = \pi.$$

Condition (i): $$\cos(2.5x) = 0.$$

We write $$2.5x = \dfrac{\pi}{2} + k\pi,$$ whence

$$x = \dfrac{\dfrac{\pi}{2} + k\pi}{2.5} = \dfrac{\pi}{5} + \dfrac{2k\pi}{5}.$$

We list integral values of $$k$$ that keep $$x$$ in $$[0,\,2\pi)$$.

• $$k = 0$$ gives $$x = \dfrac{\pi}{5}.$$

• $$k = 1$$ gives $$x = \dfrac{\pi}{5} + \dfrac{2\pi}{5} = \dfrac{3\pi}{5}.$$

• $$k = 2$$ gives $$x = \dfrac{\pi}{5} + \dfrac{4\pi}{5} = \pi.$$

• $$k = 3$$ gives $$x = \dfrac{\pi}{5} + \dfrac{6\pi}{5} = \dfrac{7\pi}{5}.$$

• $$k = 4$$ gives $$x = \dfrac{\pi}{5} + \dfrac{8\pi}{5} = \dfrac{9\pi}{5}.$$

• $$k = 5$$ gives $$x = \dfrac{\pi}{5} + \dfrac{10\pi}{5} = \dfrac{11\pi}{5} > 2\pi,$$ so we stop here.

Thus condition (i) yields the five values

$$x = \dfrac{\pi}{5},\; \dfrac{3\pi}{5},\; \pi,\; \dfrac{7\pi}{5},\; \dfrac{9\pi}{5}.$$

Collecting all distinct solutions

From (ii): $$\dfrac{\pi}{2},\; \dfrac{3\pi}{2}.$$

From (iii): $$\pi.$$

From (i): $$\dfrac{\pi}{5},\; \dfrac{3\pi}{5},\; \pi,\; \dfrac{7\pi}{5},\; \dfrac{9\pi}{5}.$$

The value $$\pi$$ appears twice, but we count it only once. Therefore the complete set, arranged in increasing order, is

$$\dfrac{\pi}{5},\; \dfrac{3\pi}{5},\; \dfrac{\pi}{2},\; \pi,\; \dfrac{7\pi}{5},\; \dfrac{3\pi}{2},\; \dfrac{9\pi}{5}.$$

There are clearly 7 distinct solutions.

Hence, the correct answer is Option A.