Question

Solve the given trigonometric equation: \[\cos x\cos 2x\cos 3x=\dfrac{1}{4}\]
(a) \[x=\left( 2n+1 \right)\dfrac{\pi }{4}\]
(b) \[x=m\pi \pm \dfrac{\pi }{6}\]
(c) \[x=m\pi \pm \dfrac{\pi }{3}\]
(d) \[x=\left( 2n+1 \right)\dfrac{\pi }{8}\]

Answer 1

Hint: Use trigonometric identity \[\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)\] to simplify the left hand side of the given equation by writing the variables ‘a’ and ‘b’ in terms of ‘x’. Simplify the equation and then equate it to the right hand side. Form equation by assuming $\cos 2x=t$ and factorize it by splitting the middle term. Solve the equation to get the values of x that satisfy the given equation.

Complete step by step answer:
We have the equation \[\cos x\cos 2x\cos 3x=\dfrac{1}{4}\]. We have to find possible values of x that satisfy the given equation. We will simplify the left hand side of the given equation and equate it to the right hand side to get the values of x which satisfy the given equation.
We know that \[\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)\].
Let’s assume \[\dfrac{a+b}{2}=3x,\dfrac{a-b}{2}=x\].
Simplifying the two equations, we have \[a+b=6x,a-b=2x\].
Adding the two equations, we get \[2a=8x\Rightarrow a=4x\].
Substituting the value \[a=4x\] in the equation \[a+b=6x\], we have \[4x+b=6x\Rightarrow b=2x\].
Thus, we can write \[\cos x\cos 3x=\dfrac{1}{2}\left( \cos 4x+\cos 2x \right)\].
So, we can write the equation \[\cos x\cos 2x\cos 3x=\dfrac{1}{4}\] as \[\cos x\cos 2x\cos 3x=\dfrac{1}{2}\left( \cos 4x+\cos 2x \right)\cos 2x=\dfrac{1}{4}\].
Simplifying the above equation, we have \[{{\cos }^{2}}2x+\cos 2x\cos 4x=\dfrac{1}{2}.....\left( 1 \right)\].
We know that \[\cos 2a=2{{\cos }^{2}}a-1\].
Substituting \[a=2x\] in the above equation, we have \[\cos 4x=2{{\cos }^{2}}2x-1\].
Substituting the above equation in equation (1), we have \[{{\cos }^{2}}2x+\cos 2x\left( 2{{\cos }^{2}}2x-1 \right)=\dfrac{1}{2}\].
Simplifying the above equation, we have \[4{{\cos }^{3}}2x+2{{\cos }^{2}}2x-2\cos 2x=1\].
Let’s assume \[\cos 2x=t\].
Thus, we have \[4{{t}^{3}}+2{{t}^{2}}-2t=1\]. We will find the roots of this equation.
We can rewrite this equation as \[2{{t}^{2}}\left( 2t+1 \right)-2t-1=0\].
Further simplifying the above equation, we have \[2{{t}^{2}}\left( 2t+1 \right)-1\left( 2t+1 \right)=0\].
Taking out the common terms, we have \[\left( 2t+1 \right)\left( 2{{t}^{2}}-1 \right)=0\].
So, we have \[t=\dfrac{-1}{2},\pm \dfrac{1}{\sqrt{2}}\].
Thus, we have \[\cos 2x=\dfrac{-1}{2},\pm \dfrac{1}{\sqrt{2}}\].
For \[\cos 2x=\dfrac{-1}{2}\], we have \[2x=\left( 2n+1 \right)\pi \pm \dfrac{\pi }{3}\Rightarrow x=\left( n+\dfrac{1}{2} \right)\pi \pm \dfrac{\pi }{6}\].
For \[\cos 2x=\pm \dfrac{1}{\sqrt{2}}\], we have \[2x=n\pi \pm \dfrac{\pi }{4}\Rightarrow x=\dfrac{n\pi }{2}\pm \dfrac{\pi }{8}\].

Hence, the solutions of the equation \[\cos x\cos 2x\cos 3x=\dfrac{1}{4}\] are \[x=\left( 2n+1 \right)\dfrac{\pi }{8}\] and \[x=m\pi \pm \dfrac{\pi }{6}\], which are options (b) and (d).

Note: One must consider all possible values of x which satisfy the given trigonometric equation. We should write the general solution of the trigonometric equation. Also, it’s important to use the trigonometric identity \[\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)\] to simplify the right hand side of the equation. One must also know the values of trigonometric functions at certain angles to find the possible solutions of the trigonometric equation.