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

Another solution could use the tangent half-angle substitution. If you define $t=\tan \frac{x}{2}$, you have $\sin x=\frac{2t}{1+t^2}$ and  $\cos x=\frac{1-t^2}{1+t^2}$. So, the equation becomes $$\cos x+8\sin x-7=\frac{16 t+2}{t^2+1}-8=0$$ that is to say $$-8 t^2+16 t-6=0$$ the roots of which being $t_1=\frac{1}{2}$ and  $t_2=\frac{3}{2}$ and so $x_1=2 \tan ^{-1}\left(\frac{1}{2}\right)$ and $x_2=2 \tan ^{-1}\left(\frac{3}{2}\right)$. 
I am sure that you can take from here.
