2018-II-38 – Solving Master

Ans: B

\begin{array}{rcl} 6 \cos^{2} x & = & \cos x + 5 \\ 6 \cos^{2} - \cos x - 5 & = & 0 \\ (6 \cos x + 5) (\cos x - 1) & = & 0 \end{array}

Therefore, $\cos x = \frac{- 5}{6}$ or $\cos x = 1$ .

For $\cos x = \frac{- 5}{6}$ , we have

$\begin{array}{rcl} \cos x & = & \frac{- 5}{6} \\ x & = & 146.442 690 2^{\circ} or 213.557 309 8^{\circ} \end{array}$

For $\cos x = 1$ , we have $x = 270^{\circ}$ .

Therefore, there are $3$ roots.