Solve the given equation $\cos x\cos 2x\cos 3x{\text{ = }}\dfrac{1}{4}$.

Hint: The equation given in the question can be solved by using the some standard formulas like $\cos (A + B) + \cos (A - B) = 2\cos A\cos B$ and $\cos 2A = {\cos ^2}A - {\sin ^2}A$ . On re-arranging the given equation and then using these formulas gives the required result. $\cos $ is positive in first and fourth quadrants (i.e., $\cos (2n\pi \pm \theta )$is always positive) $\& $ negative in second and third quadrants (i.e.,$\cos ((2n + 1)\pi \pm \theta )$ is always negative).

Formulas Used:
$\cos (A + B) + \cos (A - B) = 2\cos A\cos B$
$\cos 2A = {\cos ^2}A - {\sin ^2}A$

Complete step-by-step answer:
Given that $\cos x\cos 2x\cos 3x = \dfrac{1}{4}$
On re-arranging the given equation we get the following equation
$(2\cos 3x\cos x)(2\cos 2x) = 1$ $ \cdot \cdot \cdot \cdot \cdot (1)$
We know that $2\cos A\cos B = \cos (A + B) + \cos (A - B)$
Therefore, $2\cos 3x\cos x = \cos (3x + x) + \cos (3x - x)$
$ = \cos 4x + \cos 2x$ $ \cdot \cdot \cdot \cdot \cdot (2)$
On substituting Equation $(2)$ in Equation $(1)$
We get the following equation
$(\cos 4x + \cos 2x)(2\cos 2x) = 1$
On simplification we get the following equation
$2\cos 2x\cos 4x + 2{\cos ^2}2x = 1$
On further simplification we get the following equation
$2\cos 2x\cos 4x + (2{\cos ^2}2x - 1) = 0$ $ \cdot \cdot \cdot \cdot \cdot (3)$
We know that $\cos 2A = {\cos ^2}A - {\sin ^2}A$
We also know that ${\sin ^2}x + {\cos ^2}x = 1$
On substituting ${\sin ^2}x = 1 - {\cos ^2}x$ we get
$\cos 2A = {\cos ^2}A - (1 - {\cos ^2}A)$
$\cos 2A = 2{\cos ^2}A - 1$
Therefore, $2{\cos ^2}2x - 1 = \cos 4x$ $ \cdot \cdot \cdot \cdot \cdot (4)$
On substituting Equation $(4)$ in Equation $(3)$
We get the following equation
$2\cos 2x\cos 4x + \cos 4x = 0$
On rearranging the equation we get the following equation
$\cos 4x(2\cos 2x + 1) = 0$
In order to satisfy the above equation either anyone or both of the two terms should be equal to zero i.e.,
$\cos 4x = 0$ Or $2\cos 2x + 1 = 0$
For $\cos 4x = 0$
For $\cos 4x$ should be $0$ the value of $4x$ should be an odd multiple of $\dfrac{\pi }{2}$ .
$
  \cos 4x = \cos \dfrac{{(2n + 1)\pi }}{2} \\
    \\
 $
$4x = \dfrac{{(2n + 1)\pi }}{2}$
$x = \dfrac{{(2n + 1)\pi }}{8}$ , for all $n \in ( - \infty ,\infty )$
For $2\cos 2x + 1 = 0$
$\cos 2x = \dfrac{{ - 1}}{2}$
For $\cos 2x$ should be equal to $\dfrac{{ - 1}}{2}$ the value of $2x$ should be either in Second or Third quadrant
Therefore, $\cos 2x = \cos ((2n + 1)\pi \pm \dfrac{\pi }{3})$
$2x = (2n + 1)\pi \pm \dfrac{\pi }{3}$
$x = (2n + 1)\dfrac{\pi }{2} \pm \dfrac{\pi }{6}$ , for all $n \in ( - \infty ,\infty )$
Therefore, we get $x = (2n + 1)\dfrac{\pi }{8}$ , for all $n \in ( - \infty ,\infty )$ and $x = (2n + 1)\dfrac{\pi }{2} \pm \dfrac{\pi }{6}$ , for all $n \in ( - \infty ,\infty )$

Note: For $\cos 2x = \dfrac{{ - 1}}{2}$ i.e., for $\cos 2x$ to be negative $2x$ should be either in Second quadrant or Third quadrant since $\cos $ is negative in that quadrants. And for $\cos 4x = 0$ the value of$\cos $to be $0$ the value to $4x$ should be an odd multiple of $\dfrac{\pi }{2}$ . It is important to know the nature of $\cos $ in all four quadrants to get the required result. The above solution for $\cos 2x = \dfrac{{ - 1}}{2}$ can also be taken as $2x = 2n\pi \pm \dfrac{{2\pi }}{3}$ i.e., $x = n\pi \pm \dfrac{\pi }{3}$ , for all $n \in ( - \infty ,\infty )$ .