Explanation

(d)
Given, $3\cos{2x - 10\cos{x + 7 = 0}}$
$\Rightarrow \ \ 6\cos^{2}{x - 10\cos{x + 4 = 0}}$
$\left\lbrack \because\cos{2x = 2\cos^{2}{x - 1}} \right\rbrack$
$\Rightarrow \ \ 2\left( 3\cos{x - 2} \right)\left( \cos{x - 1} \right) = 0$
$\Rightarrow \ \ \cos{x = 1}$ or $\cos{x = \frac{2}{3}}$
Since, $\cos x$ is positive in Ist and IIIrd quadrant.
Hence, total number of solutions are 4