<s>[INST] If $8\sin x - \cos x=4$, then find possible values of $x$ I am not understanding what exactly can watch do here. First I thought that if I could square it but it was in vain. Please help me.
 [/INST] $8\sin x -\cos x =4$
solve for $\cos x$
$\cos x = 8\sin x -4$
plug in the fundamental identity $\sin^2 x + \cos^2 x=1$
$\sin^2 x + \left ( 8\sin x -4\right)^2=1$
$65 \sin ^2 x -64 \sin x +15=0$
$\sin x = \dfrac{64\pm \sqrt{64^2-4\cdot 65 \cdot 15}}{130}$
$\sin x = \dfrac{3}{5} \to x = \arcsin \left(\dfrac{3}{5}\right)+2k \pi \lor x=\pi-\arcsin \left(\dfrac{3}{5}\right)+2k \pi$
$\sin x = \dfrac{5}{13} \to x = \arcsin \left(\dfrac{5}{13}\right)+2k \pi \lor x=\pi-\arcsin \left(\dfrac{5}{13}\right)+2k \pi$
Hope this helps
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