Question

If the sum of the solutions of the equation $\cos (\frac{\pi }{3}-\theta )\cos (\frac{\pi }{3}+\theta )-\frac{\sec \theta }{4}=0$ in $[0,10\pi ]$ is $k\pi ,$ then the value of $k$ is

Answer 1

$\cos (\frac{\pi }{3}-\theta )\cos (\frac{\pi }{3}+\theta )-\frac{\sec \theta }{4}=0$
$\Rightarrow \cos \theta \cos (\frac{\pi }{3}-\theta )\cos (\frac{\pi }{3}+\theta )-\frac{1}{4}=0,\cos \theta \ne 0$
$\Rightarrow \frac{1}{4}\cos 3\theta -\frac{1}{4}=0$
$\Rightarrow \cos 3\theta =1$
$\Rightarrow 3\theta =2n\pi ,n\in Z$
$\Rightarrow \theta =\frac{2n\pi }{3}$
Since, $\theta \in [0,10\pi ],n$ should be less than or equal to $15.$
Required sum of solutions
$=\frac{2\pi }{3}\sum _{n=1}^{15}n=\frac{2\pi }{3}\times \frac{15\times 16}{2}=80\pi$