Question

Solve the following equations:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0

Answer 1

$$\begin{gathered} 3-2\cos x-4\sin x-\cos 2x+\sin 2x=0 \\ \Rightarrow 3-2\cos x-4\sin x-\left(1-2{\sin }^{2}x\right)+2\sin x\cos x=0 \\ \Rightarrow 3-2\cos x-4\sin x-1+2{\sin }^{2}x+2\sin x\cos x=0 \\ \Rightarrow \left(2{\sin }^{2}x-4\sin x+2\right)+2\cos x\left(\sin x-1\right)=0 \\ \Rightarrow 2\left({\sin }^{2}x-2\sin x+1\right)+2\cos x\left(\sin x-1\right)=0 \\ \Rightarrow 2{\left(\sin x-1\right)}^2+2\cos x\left(\sin x-1\right)=0 \\ \Rightarrow \left(\sin x-1\right)\left(2\sin x-2+2\cos x\right)=0 \\ \Rightarrow 2\left(\sin x-1\right)\left(\sin x+\cos x-1\right)=0 \\ \Rightarrow \left(\sin x-1\right)=0\mathrm{or}\left(\sin x+\cos x-1\right)=0 \\ \Rightarrow \sin x=1\mathrm{or}\sin x+\cos x=1 \\ \Rightarrow \sin x=\sin \frac{\mathrm{\pi }}{2}\mathrm{or}\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x=\frac{1}{\sqrt{2}} \\ \Rightarrow \sin x=\sin \frac{\mathrm{\pi }}{2}\mathrm{or}\sin \frac{\mathrm{\pi }}{4}\sin x+\cos \frac{\mathrm{\pi }}{4}\cos x=\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow \sin x=\sin \frac{\mathrm{\pi }}{2}\mathrm{or}\cos \left(x-\frac{\mathrm{\pi }}{4}\right)=\cos \frac{\mathrm{\pi }}{4} \\ \Rightarrow x=n\mathrm{\pi }+{\left(-1\right)}^n\frac{\mathrm{\pi }}{2}\mathrm{or}x-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4},n\in \mathrm{\mathbb{Z}} \\ \Rightarrow x=n\mathrm{\pi }+{\left(-1\right)}^n\frac{\mathrm{\pi }}{2}\mathrm{or}x=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2}\mathrm{or}x=2n\mathrm{\pi },n\in \mathrm{\mathbb{Z}} \end{gathered}$$