algebra precalculus - Solve $cos x+8sin x-7=0$

Solve $\cos x+8\sin x-7=0$

My attempt:

$$\begin{align} &8\sin x=7-\cos x\\ &\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\ &\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\ &\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\ &\implies 8\sin \frac{x}{2}\cos \frac{x}{2}=3+\sin^2 \frac{x}{2}\\ &\implies 0=\sin^2 \frac{x}{2}-8\sin \frac{x}{2}\cos \frac{x}{2}+3\\ &\implies 0=\sin \frac{x}{2}\left(\sin \frac{x}{2}-8\cos \frac{x}{2}\right)+3 \end{align}$$

I'm not sure how to proceed from here (if this process is even right at all?) . Any help would be appreciated. Thanks.

Answer

Divide either sides by $\cos^2\frac x2$ $$\sin^2\frac x2-8\sin\frac x2\cos\frac x2+3=0$$ to get $$\tan^2\frac x2-8\tan\frac x2+3\left(1+\tan^2\frac x2\right)=0$$ which on rearrangement is a Quadratic Equation in $\displaystyle\tan\frac x2$