Find the general solution of each of the following equations:

To Find: General solution.

Given: $4 \sin x \cos x+2 \sin x+2 \cos x+1=0 \Rightarrow 2 \sin x(2 \cos x+1)+2 \cos x+1=0$

So $(2 \cos x+1)(2 \sin x+1)=0$

$\cos x=\frac{-1}{2}=\cos \left(\frac{2 \pi}{3}\right)$ or $\sin x=\frac{-1}{2}=\sin \frac{7 \pi}{6}$

Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 \mathrm{n} \pi \pm \alpha$ or $\sin \theta=\sin \alpha \Rightarrow \theta=\mathrm{m} \pi+(-1)^{\mathrm{m}} \alpha$ where $n, m \in I$

$x=2 n \pi \pm \frac{2 \pi}{3}$ or $x=m \pi+(-1)^{m} \cdot \frac{7 \pi}{6}$ where $n, m \in \mid$

So the general solution is $x=2 n \pi \pm \frac{2 \pi}{3}$ or $x=m \pi+(-1)^{m} \cdot \frac{7 \pi}{6}$ where $n, m \in I$