Question

$\mathrm{Find}\mathrm{the}\mathrm{possible}\mathrm{values}\mathrm{of}\sin x\mathrm{if}8\sin x-\cos x=4$

Answer 1

$$\begin{gathered} \mathrm{Given}:8\sin x-\cos x=4 \\ \Rightarrow 8\sin x-4=\cos x \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ 64{\sin }^{2}x-64\sin x+16={\cos }^{2}x \\ \Rightarrow 64{\sin }^{2}x-64\sin x+16=1-{\sin }^{2}x \\ \Rightarrow 65{\sin }^{2}x-64\sin x+15=0 \\ \Rightarrow 65{\sin }^{2}x-39\sin x-25\sin x+15=0 \\ \Rightarrow 13\sin x(5\sin x-3)-5(5\sin x-3)=0 \\ \Rightarrow (5\sin x-3)(13\sin x-5)=0 \\ \therefore \sin x=\frac{3}{5},\frac{5}{13} \end{gathered}$$