<s>[INST] How to solve $\sin(x) + 2\sqrt{2}\cos x =3$ How to solve $\sin(x) + 2\sqrt{2}\cos x=3$ ?
What is general method for doing these kind of questions?
Thanks
 [/INST] Your problem is of the following form 
$$a\sin x+b\cos x = c$$
where $a = 1$, $b = 2\sqrt{2}$ and $c = 3$.
Let $R = \sqrt{a^2 + b^2}$. We can define $$A=\dfrac{a}{R} =\cos\theta$$ and $$B=\dfrac{b}{R} =\sin\theta$$
Therefore
$$\begin{align*}
a\sin x+b\cos x&=R(A\sin x+B\cos x)=R(\cos\theta\sin x+\sin\theta\cos x)=R\sin(x+\theta)\;.
\end{align*}$$
Hence $$\sin(x+\theta)=\frac{c}{R}\;$$ 
or
$$x=(\sin^{-1}\frac{c}{R})- \theta = (\sin^{-1}\frac{c}{R})- (\sin^{-1}\frac{b}{R}) = (\sin^{-1}\frac{3}{3})- (\sin^{-1}\frac{2\sqrt{2}}{3}) = 0.3398$$ 
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