Question

The arithmetic mean of the roots of the equation
$4{\cos }^{2}x-4{\cos }^{2}x-\cos (315\pi +x)=1$ in the interval $(0,315)$ is

• A. $50\pi$
• B. $51\pi$
• C. $100\pi$
• D. $315\pi$

Answer 1

The correct option is B $51\pi$

Given
$4{\cos }^{3}x-4{\cos }^{3}x-\cos (315\pi +x)=1$
$\Rightarrow 4{\cos }^{3}x-4{\cos }^{3}x-\cos x-1=0[\because \cos (315\pi +x)={(-1)}^{315}\cos x=-\cos x]$
$\Rightarrow (4{\cos }^{2}x+1)(\cos x-1)=0$
$\Rightarrow \cos x-1=0$
$4{\cos }^{2}x+1\ne 0$
$\Rightarrow \cos x=1$
$\Rightarrow \cos x=\cos 0$
$\Rightarrow x=2n\pi ,n\in I$
$\therefore x=2\pi ,4\pi ,6\pi ,8\pi ,...100\pi [\because 0<x<315]$
$\therefore$ Required Arithmetic mean
$=\frac{2\pi +4\pi +6\pi +8\pi ,...+100\pi }{50}=\frac{2\pi .\frac{50}{2}.51}{50}=51\pi$