Ans: (b) $\dfrac{8\pi}{9}$ or $\dfrac{10\pi}{9}$
-
$\begin{array}{cl}
& \cos 3x \\
= & \cos (2x+x) \\
= & \cos 2x \cos x -\sin 2x \sin x \\
= & (2\cos^2 x-1)\cos x -2\sin x \cos x \sin x \\
= & 2\cos^3 x -\cos x -2\sin^2 x \cos x \\
= & 2\cos^3 x -\cos x -2(1-\cos^2 x)\cos x \\
= & 2\cos^3 x-\cos x-2\cos x+2\cos^3 x \\
= & 4\cos^3 x-3\cos x
\end{array}$ -
$\begin{array}{rcl}
\sec^3 x-6\sec^2 x+8 & = & 0 \\
\dfrac{1}{\cos^3 x}-\dfrac{6}{\cos^2 x}+8 & = & 0 \\
1-6\cos x+8\cos^3 x& = & 0 \\
2(4\cos^3 x-3\cos x) & = & -1 \\
\cos 3x & = & \dfrac{-1}{2} \\
\end{array}$$\therefore 3x =\pi-\dfrac{\pi}{3}$ or $3x=\pi+\dfrac{\pi}{3}$ or $3x=3\pi-\dfrac{\pi}{3}$ or $3x=3\pi+\dfrac{\pi}{3}$.
$\therefore x =\dfrac{2\pi}{9}$ (rejected) or $x=\dfrac{4\pi}{9}$ (rejected) or $x=\dfrac{8\pi}{9}$ or $x=\dfrac{10\pi}{9}$.
