Question

The sum of values of $x$ satisfying $\cos 4x+6=7\cos 2x$ in the interval $[0,314]$ is $k\pi ,k\in R,$ then $k$ is equal to

• A. $4950$
• B. $5050$
• C. $5000$
• D. None of these

Answer 1

The correct option is B $5050$

$\cos 4x+6=7\cos 2x$
$\Rightarrow 2{\cos }^{2}2x-1+6=7\cos 2x$
$\Rightarrow 2{\cos }^{2}2x-7\cos 2x+5=0$
$\Rightarrow 2{\cos }^{2}2x-5\cos 2x-2\cos 2x+5=0$
$\Rightarrow 2\cos 2x(\cos 2x-1)-5(\cos 2x-1)=0$
$\Rightarrow cos2x=1$ or $\cos 2x=\frac{5}{2}$
$\Rightarrow \cos 2x=\frac{5}{2}$ is not possible.
Hence, $\cos 2x=1$

$2x=2n\pi \pm 0$

$2x=0,2\pi ,4\pi ...200\pi$

$x=0,\pi ,...100\pi$

Sum of values of $x,k\pi =0+\pi +...+100\pi$

$k\pi =\pi (1+2+3+.........100)$

$k\pi =\pi \left(\frac{100\times 101}{2}\right)$

$k\pi =5050\pi$

Thus, $k=5050$