The sum of the roots of the equation $\cos 4x + 6 = 7\cos 2x$ in the interval $[0,314]$ is $\lambda r$, then the numerical value of $\Lambda$ is
Step-by-Step Solution
Key Concept: Convert trigonometric equations to polynomial form using substitution, then solve for the parameter and back-substitute to find all solutions in the given interval.
Starting with $(2\cos^2 2x - 1) - 6 = 7\cos 2x$, we substitute $t = \cos 2x$ to get $2t^2 - 7t - 6 = 0$. Factoring: $(2t + 1)(t - 6) = 0$ gives $t = -\frac{1}{2}$ or $t = 6$ (impossible since $|\cos 2x| \leq 1$). From $\cos 2x = -\frac{1}{2}$, we get $2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \ldots$, so $x = n\pi + \frac{\pi}{3}$ for integer $n$ (accounting for both positive and negative angles in $[0, 2\pi)$). The roots in $[0, 314]$ are $\pi, 2\pi, 3\pi, \ldots, 99\pi$ plus fractional parts. The sum equals $99 \cdot 50 = 4950$.
Correct Answer: 4950