Question

Total number of solutions of $\cos x\cos 2x\cos 3x=\frac{1}{4}$ in $[0,\pi ]$ is equal to

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

The correct option is B 6

$\begin{array}{c}\cos x\cos 2x\cos 3x=\frac{1}{4}\\ \left(2\cos x\cos 3x\right)\cos 2x=\frac{1}{2}\\ \left(\cos 2x+\cos 4x\right)\cos 2x=\frac{1}{2}\\ Let\cos 2x=2\left[\cos 4x=2{\cos }^{2}2x-1\right]\\ Now,\\ \left(t+2t^2-1\right)t=\frac{1}{2}\\ 2t\left(2t^2+t-1\right)=1\\ 4t^3+2t^2-2t-1=0\\ 2t^2\left(2t+1\right)-\left(2t-1\right)=0\\ \left(2t^2-1\right)\left(2t+1\right)=0\\ So,t=\frac{-1}{2}t=\pm \frac{1}{\sqrt{2}}\\ From{}^{\prime }{t}^{\prime }=\frac{-1}{2}\\ \cos 2x=-\frac{1}{2}\\ \Rightarrow 2x=\frac{2\pi }{3},\frac{4\pi }{3}\\ Andfromt=\pm \frac{1}{\sqrt{2}}\\ \cos 2x=\frac{1}{\sqrt{2}}\Rightarrow 2x=\frac{\pi }{4},\frac{7\pi }{4}\\ x=\frac{\pi }{8},\frac{7\pi }{8}\\ Now,\cos 2x=\frac{-1}{\sqrt{2}}\\ 2x=\frac{3\pi }{4},\frac{5\pi }{4}\\ \Rightarrow \frac{\pi }{8},\frac{\pi }{3},\frac{3\pi }{8},\frac{5\pi }{8},\frac{2\pi }{3},\frac{7\pi }{8}\\ thereare6solutions.\\ Hence,optionBiscorrectanswer.\end{array}$