Question

The arithmetic mean of the roots of the equation$4{\cos }^{3}x-4{\cos }^{2}x-\cos (315𝜋+x)=1$ in the interval $(0,315)$ is equal to?

Answer 1

We know that$\cos \left[\right(2n+1)\mathrm{\pi }+x]=–\cos xn\in I$

So, $\cos \left(315\mathrm{\pi }+\mathrm{x}\right)=-\cos x$

Therefore, the given equation becomes

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

$\begin{array}{rcl}& \Rightarrow & 4{\cos }^{2}x(\cos x–1)+\left(\cos x–1\right)=0\\ & \Rightarrow & (4{\cos }^{2}x+1)(\cos x–1)=0\end{array}$

$\cos x=1$or${\cos }^{2}x=-\frac{1}{4}$(not possible)

So,

$\begin{array}{rcl}\cos x& =& 1\\ & \Rightarrow & \cos x=\cos 0°\\ & \Rightarrow & x=2n\mathrm{\pi }n\in I\end{array}$

Now, we see that $100\mathrm{\pi }<315<101\mathrm{\pi }$

$\therefore x=2\pi ,4\pi ,6\pi ,8\pi ,...100\pi [\because 0<x<315]$

We have , there are $50$ terms which are in A.P.

Sum of $50$terms,

$\begin{array}{rcl}S& =& \frac{50}{2}\times (2\mathrm{\pi }+100\mathrm{\pi })\\ & =& 25\times 102\mathrm{\pi }\end{array}$

Thus, Arithmetic mean $=\frac{25+102\mathrm{\pi }}{50}=51\mathrm{\pi }$

Hence, The arithmetic mean of roots of the given equation is $51\mathrm{\pi }$