<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] Another way to solve $A \sin\theta + B \cos\theta = C$ is as follows.
Divide by $\sqrt{A^2 + B^2}$.  Then, set $A / \sqrt{A^2 + B^2} = \sin\xi$
and $B / \sqrt{A^2 + B^2} = \cos\xi$.
This gives:
$$\sin\xi \sin\theta + \cos\xi \cos\theta = C / \sqrt{A^2 + B^2}.$$
So, the solution is given by
$$\cos(\theta - \xi) = C / \sqrt{A^2 + B^2},$$
where $\xi$ is the angle that the vector $(A,B)$ makes with the positive $x$ semi-axis.
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