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

If you make the following substitutions:
$$
\begin{cases}
X=\cos x\\
Y=\sin x
\end{cases}
$$
then your equations (remembering that $\sin^2 x+\cos^2x=1$) becomes
$$
\begin{cases}
X^2+Y^2=1\\
Y+2\sqrt{2}X=3
\end{cases}
$$
which is the intersection between a circle and a straight line.
Solving it:
$$
\begin{cases}
Y=3-2\sqrt{2}X\\
X^2+(3-2\sqrt{2}X)^2=1\\
\end{cases}
$$
you obtain the equation
$$
X^2+9-12\sqrt{2}X+8X^2=1,
$$
which is
$$
\begin{align*}
9X^2-12\sqrt{2}X+8&=0,\\
(3X-2\sqrt{2})^2&=0,
\end{align*}
$$
which yelds the solution
$$
\begin{cases}
X=\dfrac{2\sqrt{2}}{3}\\
Y=3-2\sqrt{2}\dfrac{2\sqrt{2}}{3}=3-\dfrac{8}{3}=\dfrac{1}{3}
\end{cases}
$$
Substituting back:
$$
\begin{cases}
\cos x=\dfrac{2\sqrt{2}}{3}\\
\sin x=\dfrac{1}{3}
\end{cases}
$$
Or:
$$
x=\arctan\dfrac{1}{2\sqrt2}+2k\pi,\quad\text{for $k\in\mathbf{Z}$}.
$$
