-
$\begin{array}{cl}
& 4\cos x\cos 2x \cos 3x -1 \\
= & 2(2\cos 2x \cos x)\cos 3x -1 \\
= & 2(\cos x+\cos 3x)\cos 3x -1 \\
= & 2\cos x \cos 3x+2\cos^23x -1 \\
= & 2\cos 3x \cos x + 2\cos^2 3x -1 \\
= & \cos 2x+\cos 4x+\cos 6x
\end{array}$ - By (a), we have
$\begin{array}{rcl}
\cos4\theta +\cos 8\theta +\cos 12\theta & = & -1 \\
4\cos 2\theta\cos 4\theta\cos 6\theta-1 & = & -1 \\
\cos 2\theta \cos 4\theta \cos 6\theta & = & 0
\end{array}$$\therefore \cos 2\theta =0$ or $\cos 4\theta =0$ or $\cos 6\theta =0$.
$\therefore 2\theta =\dfrac{\pi}{2}$ or $2\theta=\dfrac{3\pi}{2}$ or $4\theta =\dfrac{\pi}{2}$ or $4\theta=\dfrac{3\pi}{2}$ or $4\theta=\dfrac{5\pi}{2}$ or $6\theta =\dfrac{\pi}{2}$ or $6\theta=\dfrac{3\pi}{2}$ or $6\theta=\dfrac{5\pi}{2}$ or $6\theta=\dfrac{7\pi}{2}$.
$\therefore \theta=\dfrac{\pi}{4}$ or $\dfrac{3\pi}{4}$ (rejected) or $\theta=\dfrac{\pi}{8}$ or $\theta=\dfrac{3\pi}{8}$ or $\theta =\dfrac{5\pi}{8}$ (rejected) or $\theta=\dfrac{\pi}{12}$ or $\theta=\dfrac{\pi}{4}$ or $\theta=\dfrac{5\pi}{12}$ or $\theta=\dfrac{7\pi}{12}$ (rejected).
$\therefore \theta =\dfrac{\pi}{12}$ or $\theta =\dfrac{\pi}{8}$ or $\theta =\dfrac{\pi}{4}$ or $\theta =\dfrac{3\pi}{8}$ or $\theta =\dfrac{5\pi}{12}$.
2021-M2-04
Ans: (b) $\dfrac{\pi}{12}$ or $\dfrac{\pi}{8}$ or $\dfrac{\pi}{4}$ or $\dfrac{3\pi}{8}$ or $\dfrac{5\pi}{12}$
