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.
 A: 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.
A: 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 $\displaystyle\tan\frac x2$
A: 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.
A: $1.) \;  \cos x+8\sin x-7=0; \tag{1}$
$2.) \; \cos^2 x = (7- 8\sin x)^2; \tag{2}$
$3.) \; \cos^2 x = 1 - \sin^2 x; \tag{3}$
$4.) \; 1 - \sin^2 x = (7- 8\sin x)^2; \tag{4}$
$5.)  \; 1 - \sin^2 x = 49 - 112 \sin x + 64 \sin^2 x; \tag{5}$
$6.) \; 65 \sin^2 x -112\sin x + 48 = 0; \tag{6}$
$7.) \; \sin x = \dfrac{112 \pm \sqrt{(112)^2 - 4(48)(65)}}{130}; \tag{7}$
$8.) \; \sin x = \dfrac{112 \pm \sqrt{12,544 -12,480}}{130}; \tag{8}$
$9.) \sin x = \dfrac{112 \pm 8}{130} = \dfrac{104}{130}, \dfrac{120}{130} = \dfrac{4}{5}, \dfrac{12}{13}; \tag{9}$
$10.) \; \text{Note there exist unique} \; \alpha, \beta \in [0, \dfrac{\pi}{2}] \; \text{with} \; \sin \alpha = \dfrac{4}{5}, \sin \beta = \dfrac{12}{13}; \tag{10}$
$11.) \; \text{Note} \; \cos \alpha = \dfrac{3}{5}, \cos \beta = \dfrac{5}{13}; \tag{11}$
$12.) \; \text{Note} \; \alpha \; \text{alone satisfies (1)}; \tag{12}$
$13.) \; \text{Note there exist unique} \; \gamma, \delta \in [\dfrac{\pi}{2}, \pi] \; \text{with} \; \sin \gamma = \dfrac{4}{5}, \sin \delta = \dfrac{12}{13}; \tag{13}$
$14.) \; \text{Note that} \; \cos \gamma = -\dfrac{3}{5}, \cos \delta = -\dfrac{5}{13}; \tag{14}$
$15.) \; \text{Note that} \; \delta \; \text{alone satisfies (1)}; \tag{15}$
$16.) \; \text{Note by periodicity} \; \alpha \pm 2n\pi, \delta \pm 2n\pi \; \text{satisfy (1) for} \; n \in \Bbb Z; \tag{16}$
$17.) \; \text{Conclude that all solutions are given by item (16) above}; \tag{17}$
$18.) \; \text{Kick back for a minute; you deserve it!} \tag{18}$
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
