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

The correct option is D 4

Given, $3\cos 2x–10\cos x+7=0$
$\Rightarrow 3(2{\cos }^{2}x-1)-10\cos x+7=0$
$\Rightarrow 6{\cos }^{2}x-10\cos x+4=0$
$\Rightarrow 3{\cos }^{2}x-5\cos x+2=0$
$\Rightarrow (3\cos x-2)(\cos x-1)=0$
$\Rightarrow \cos x=1or\cos x=\frac{2}{3}$
$\Rightarrow x=0,2\pi orx={\cos }^{-1}\frac{2}{3},2\pi -{\cos }^{-1}\frac{2}{3}$
Hence, four values exit, when $x\in [0,2\pi ].$