Question

Find the possible values of x if \[\cos x.\cos 2x.\cos 3x=\dfrac{1}{4}\].

Answer 1

Hint: Multiply by 4 in LHS and RHS of the equation \[\cos x.\cos 2x.\cos 3x=\dfrac{1}{4}\] . Now, transform the equation using the formula \[2.\cos A.\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\] . Consider A as x and B as 3x. Then, replace \[2{{\cos }^{2}}2x-1\] by \[\cos 4x\]. We know the general solution of \[\cos \theta =0\] is \[\theta =(2n+1)\dfrac{\pi }{2}\] and \[\cos \alpha =-\dfrac{1}{2}\] is \[\alpha =2n\pi \pm \dfrac{2\pi }{3}\]. Solve it further using these general solutions.

Complete step-by-step solution -
According to the question we have the equation, \[\cos x.\cos 2x.\cos 3x=\dfrac{1}{4}\] …………….(1)
Now, multiplying by 4 in LHS and RHS of the equation (1), we get
\[\cos x.\cos 2x.\cos 3x=\dfrac{1}{4}\]
\[\Rightarrow 4.\cos x.\cos 2x.\cos 3x=1\]
\[\Rightarrow 2.\cos x.\cos 3x.(2\cos 2x)=1\] ……………………..(2)
We have to simplify equation (2).
We know the formula, \[2.\cos A.\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\] …………(3)
Replacing A by x and B by 3x in equation (3), we get
\[2.\cos x.\cos 3x=\cos \left( x+3x \right)+\cos \left( x-3x \right)\]
\[\Rightarrow 2.\cos x.\cos 3x=\cos \left( x+3x \right)+\cos \left( -2x \right)\] …………………(4)
We know that, \[\cos (-x)=\cos x\]. Replacing x by 2x, we get
\[\cos (-2x)=\cos 2x\] ……………….(5)
Now, using (5), transforming equation (4),
\[\Rightarrow 2.\cos x.\cos 3x=\cos \left( 4x \right)+\cos \left( 2x \right)\] ……………….(6)
Now, using (6), transforming equation (2),
\[\{\cos \left( 4x \right)+\cos \left( 2x \right)\}2\cos 2x=1\]
\[\Rightarrow 2\cos 4x.\cos 2x+2{{\cos }^{2}}2x=1\]
\[\Rightarrow 2\cos 4x.\cos 2x+2{{\cos }^{2}}2x-1=0\] ………………….(7)
We know the formula, \[2{{\cos }^{2}}x-1=\cos 2x\] .
Now, replacing x by 2x in the above formula, we get
\[2{{\cos }^{2}}2x-1=\cos 4x\] ……………..(8)
Now, using (8), transforming equation (7), we get
\[\begin{align}
  & \Rightarrow 2\cos 4x.\cos 2x+\cos 4x=0 \\
 & \Rightarrow \cos 4x(2\cos 2x+1)=0 \\
\end{align}\]
\[\cos 4x=0\] or \[2\cos 2x+1=0\] .
We know that the general solution of \[\cos \theta =0\] is \[\theta =(2n+1)\dfrac{\pi }{2}\] .
\[\cos 4x=0\]
Now, using the general solution of \[\cos \theta =0\] , we get
\[\begin{align}
  & \Rightarrow 4x=(2n+1)\dfrac{\pi }{2} \\
 & \Rightarrow x=(2n+1)\dfrac{\pi }{8} \\
\end{align}\]
\[\begin{align}
  & 2\cos 2x+1=0 \\
 & \Rightarrow 2\cos 2x=-1 \\
 & \Rightarrow \cos 2x=\dfrac{-1}{2} \\
\end{align}\]
We know that, \[\cos \left( \pi -\dfrac{\pi }{3} \right)=\cos \left( \dfrac{2\pi }{3} \right)=-\dfrac{1}{2}\]
We have to find the general solution of \[\cos \theta =-\dfrac{1}{2}\] .
\[\begin{align}
  & \cos \theta =\cos \left( \pi -\dfrac{\pi }{3} \right) \\
 & \Rightarrow cos\theta =\cos \left( \dfrac{2\pi }{3} \right) \\
\end{align}\]
\[\Rightarrow \theta =2n\pi \pm \dfrac{2\pi }{3}\]
Now, we have the general solution of \[\cos \theta =-\dfrac{1}{2}\] equal to \[\theta =2n\pi \pm \dfrac{2\pi }{3}\] .
Now, using the general solution of \[\cos \theta =-\dfrac{1}{2}\] , we get
\[\begin{align}
  & \Rightarrow 2\cos 2x=-1 \\
 & \Rightarrow \cos 2x=\dfrac{-1}{2} \\
\end{align}\]
Replacing 2x by \[\theta \] in the above equation, we get
\[\cos \theta =-\dfrac{1}{2}\]
we have the general solution of \[\cos \theta =-\dfrac{1}{2}\] equal to \[\theta =2n\pi \pm \dfrac{2\pi }{3}\]……..(9)
Now, replacing \[\theta \] by 2x in equation (9), we get
\[\begin{align}
  & \Rightarrow 2x=2n\pi \pm \dfrac{2\pi }{3} \\
 & \Rightarrow x=n\pi \pm \dfrac{\pi }{3} \\
\end{align}\]
So, \[x=n\pi \pm \dfrac{\pi }{3}\] or \[x=(2n+1)\dfrac{\pi }{8}\].

Note: In this question, one can consider 2x as A and 3x as B and then transform the equation \[4\cos x.\cos 2x.\cos 3x=1\] using the formula \[2.\cos A.\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\] . If we do so, then we will get one 5x term due to which our equation will become complex to solve. So, this approach is wrong for this question. Therefore, we have to consider x as A and 3x as B and then transform \[4\cos x.\cos 2x.\cos 3x=1\] the equation using the formula \[2.\cos A.\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\] .