# The number of values of $x$ in $[0,2\pi]$ satisfying the equation $3\cos 2x-10\cos x+7=0$ is?

$(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;4$

Given :
$3\cos 2x-10\cos x+7=0$
$\Rightarrow\:3(2\cos^2x-1)-10\cos x+7=0$
$\Rightarrow\:6\cos^2x-10\cos x+4=0$
$\Rightarrow\:3\cos^2x-5\cos x+2=0$
$\Rightarrow\:(3\cos x-2)(\cos x-1)=0$
$\Rightarrow\:\cos x=1$ or $cos\:x=\large\frac{2}{3}$
$\Rightarrow\:x=0,2\pi$ or
$\Rightarrow\:x=\cos^{-1}\large\frac{2}{3}$
or $x=2\pi-\cos^{-1}\large\frac{2}{3}$
Hence we have 4 solutions.
$(d)$ is the correct answer.
edited Mar 25, 2014