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

& \Rightarrow 2\cos 4x.\cos 2x+\cos 4x=0 \\\ & \Rightarrow \cos 4x(2\cos 2x+1)=0 \\\ \end{aligned}$$ $$\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{aligned} & \Rightarrow 4x=(2n+1)\dfrac{\pi }{2} \\\ & \Rightarrow x=(2n+1)\dfrac{\pi }{8} \\\ \end{aligned}$$ $$\begin{aligned} & 2\cos 2x+1=0 \\\ & \Rightarrow 2\cos 2x=-1 \\\ & \Rightarrow \cos 2x=\dfrac{-1}{2} \\\ \end{aligned}$$ We know that, $$\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{aligned} & \cos \theta =\cos \left( \pi -\dfrac{\pi }{3} \right) \\\ & \Rightarrow cos\theta =\cos \left( \dfrac{2\pi }{3} \right) \\\ \end{aligned}$$ $$\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{aligned} & \Rightarrow 2\cos 2x=-1 \\\ & \Rightarrow \cos 2x=\dfrac{-1}{2} \\\ \end{aligned}$$ Replacing 2x by $$\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{aligned} & \Rightarrow 2x=2n\pi \pm \dfrac{2\pi }{3} \\\ & \Rightarrow x=n\pi \pm \dfrac{\pi }{3} \\\ \end{aligned}$$ So, $$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)$$ .