Question

The sum of the roots of the equation $4{\cos }^3(\pi +x)-4{\cos }^2(\pi -x)+\cos (\pi +x)-1=0$ in the interval $[0,320]$ is p$\pi$ where p is equal to

• A. $2500$
• B. $2601$
• C. $2600$
• D. $2651$

Answer 1

The correct option is B $2601$

Simplifying the equation, we get
$-4{\cos }^{3}x-4{\cos }^{2}x-\cos \left(x\right)-1=0$
Or
$4{\cos }^{3}x+4{\cos }^{2}x+\cos \left(x\right)+1=0$
Or
$\left[4{\cos }^2\right(x)+1]\left[\cos \right(x)+1]=0$
Now
${\cos }^2\left(x\right)=\frac{-1}{4}$ is not possible.
Hence
$\cos \left(x\right)=-1$
Or $x=(2n-1)\pi$ where $nϵN$.
Hence
$x=\pi ,3\pi ,5\pi ...$
Now
$101\pi <320<102\pi$.
Hence
$x=\pi ,3\pi ,5\pi ...101\pi$.
Now
$Sum$
$=\pi [1+3+5+...101]$
$=\pi \left[\right(1+2+3+..101)-(2+4+6+8+...100\left)\right]$
$=\pi [\frac{101.102}{2}-2(\frac{50.51}{2}\left)\right]$

$=\pi \left[51\right(101)-51(50\left)\right]$

$=\pi \left[51\right(51\left)\right]$
$=\pi .2601.$