The sum of the roots of the equation 4x + 6 = 7 2x in the interval...

$\left(2\cos^{2} 2x - 1\right) + 6 = 7\cos 2x$
On putting $\cos 2x = t$, we get,
$2t^{2} - 1 + 6 = 7t$
$2t^{2} - 7t + 5 = 0$
\left(2t - 5\right)\left(t - 1\right) = 0$<br />$t = \dfrac{5}{2}, 1$<br />$t = \dfrac{5}{2}$~(not possible)<br />$t = 1 ⇒ \cos 2x = 1 ⇒ 2x = 2n\pi$<br />$⇒ x = n\pi$<br />The roots in~\left[0,314\right]$ are
$\pi, 2\pi, 3\pi, . . . . . , 99\pi \left\{100\pi > 314\right\}<br />Sum$ of roots= \pi + 2\pi + 3\pi + . . . . + 99\pi = 4950\pi$<br />$⇒ \lambda = 4950