Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

Given, $3\cos 2x-10\cos x+7=0$
$\Rightarrow 3(2{\cos }^{2}x-1)-10\cos x+7=0$
$\Rightarrow 6{\cos }^{2}x-10\cos x+4=0$

$\Rightarrow 2(3{\cos }^{2}x-5\cos x+2)=0$
$\Rightarrow 2\left[(3\cos x-2)(\cos x-1)\right]=0$

$\Rightarrow (3\cos x-2)(\cos x-1)=0$
$\Rightarrow \cos x=1$ and $\cos x=\frac{2}{3}$
Since, $\cos x$ is positive in Ist and IVth quadrants.
Hence, total number of solution is $4$.