Question

Solve $(cos\frac{\theta }{4}-2sin\theta )sin\theta +(1+sin\frac{\theta }{4}-2cos\theta )cos\theta =0$

Answer 1

$(\frac{\cos \theta }{4}-2\sin \theta )\sin \theta +(1+\frac{\sin \theta }{4}-2\cos \theta )\cos \theta =0$

$\sin \theta \cos \frac{\theta }{4}-2{\sin }^2\theta +\cos \theta +\sin \frac{\theta }{4}\cos \theta -2{\cos }^2\theta =0$

$\sin \theta \frac{\theta }{4}+\sin \frac{\theta }{4}\cos \theta +\cos \theta -2({\sin }^2\theta +{\cos }^2\theta )=0$

$\downarrow$

$\sin a\cos b+\cos a\sin b$

$\sin (\theta +\frac{\theta }{4})+\cos \theta -2=0[{\sin }^2\theta +{\cos }^2\theta =1]$

$\sin \frac{5\theta }{4}+\cos \theta =2$

For this to happen as we know $\sin \frac{5\theta }{4}\le 1<\cos \theta \le 1$

Thus $\sin \frac{5\theta }{4}+\cos \theta \le 2$. Therefore equality exists

Thus $\sin \frac{5\theta }{4}=1$ & $\cos \theta =1$

$\frac{5\theta }{4}=(4n_1+1)\frac{\pi }{2}{n}_1\in Z\theta =2n_2\pi n_2\in Z$

$\theta =(4n_1+1)\frac{2\pi }{5}{n}_1\in Z.....\left(1\right)\theta =2n_2\pi ...\left(2\right)n_2\in Z$

Taking intersection of $\left(1\right)$ & $\left(2\right)$

$\theta =(4n+10)2\pi n\in Z$

$\theta =\{2\pi ,10\pi ,18\pi ,.....\}$