The sum of the roots of the equation  $4{\cos }^{3}x-4{\cos }^{2}x-\cos \left(\pi +x\right)-1=0$ in the interval $\left[0,315\right]$ is $p\pi$ , where p is equal to

We have $4{\cos }^{3}x-4{\cos }^{2}x-\cos \left(\pi +x\right)-1=0$

            $\Rightarrow 4{\cos }^{3}x-4{\cos }^{2}x+\cos x-1=0$

            $\Rightarrow \left(4{\cos }^{2}x+1\right)\left(\cos x-1\right)=0$

             $\Rightarrow \cos x=1\Rightarrow x=2n\pi$

              $Wehave100\pi <315<101\pi$

$\therefore$Required sum = $2\pi +4\pi +\mathrm{......100}\pi$

                                   $=2\left(1+2+\mathrm{....50}\right)\pi$

                                    $\begin{array}{l}=\frac{2\times 50\times 51}{2}\pi \\ =2550\pi \end{array}$