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.

Use the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi) $$ where $C=\sqrt{A^2+B^2},\phi=\arg (A+Bi)$.
This makes transforms your equation to:
$$\sqrt{65} \sin(x+\tan^{-1} 8)=7 $$
which can be easily solved.
