Question

Find the number of all values of $\theta \in [0,10.5]$ satisfying the equation $\cos 6+\cos 4\theta +\cos 2\theta +1=0$.

Answer 1

Given:-$f_n\left(\theta \right)=\tan \left(\frac{\theta }{2}\right)(1+\sec \theta )(1+\sec 2\theta )(1+\sec {}^{2}2\theta ).......(1+\sec 2^n\theta )$

$f_0\left(\theta \right)=\tan \left(\frac{\theta }{2}\right)(1+\sec \theta )=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\frac{2\cos {}^2\frac{\theta }{2}}{\cos \theta }=\tan \theta f_1\left(\theta \right)=\tan \theta \frac{2\cos {}^2\theta }{\cos 2\theta }=\tan 2\theta f_2\left(\theta \right)=\tan 2\theta \frac{2\cos {}^{2}4\theta }{\cos 4\theta }=\tan 4\theta .....f_n\left(\theta \right)=\tan 2^n\theta \therefore f_2\left(\frac{\pi }{16}\right)=\tan \frac{\pi }{4}=1f_3\left(\frac{\pi }{32}\right)=\tan \frac{\pi }{4}=1f_4\left(\frac{\pi }{64}\right)=\tan \frac{\pi }{4}=1f_5\left(\frac{\pi }{128}\right)=\tan \frac{\pi }{4}=1$