Question

Solve:

$3-2\cos \theta -4\sin \theta -\cos 2\theta +\sin 2\theta =0$.

Answer 1

$$\begin{array}{c}2+(1-\cos 2\theta )-4\sin \theta +2\sin \theta .\cos \theta -2\cos \theta =0\\ 2-2\sin \theta +2\sin \theta +2{\sin }^2\theta +2\sin \theta .\cos \theta -2\cos \theta =0\\ -2(\sin \theta -1)+2\sin \theta (\sin \theta -1)+2\cos (2\sin \theta -1)=0\\ =2(\sin \theta -1)[\sin 2\theta +\cos \theta -1]=0\\ \sin \theta -1=0\\ \sin \theta =1\\ \theta =n\pi +(-1{)}^{n}1,n\in I\\ or\\ Sin\theta +\cos \theta =1\\ \cos (\theta -\frac{\pi }{4})=\frac{1}{\sqrt{2}}\\ \cos (\theta -\frac{\pi }{4})=\cos \pi 14\\ \therefore \theta =2k\pi \pm \frac{\pi }{4}+\frac{\pi }{4},K\in I\end{array}$$