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
 A: Geometrically, the equation says that the dot product $\vec v\cdot(2\sqrt2,1)$ is $3$, where $\vec v$ is the unit vector $(\cos x,\sin x)$.
Since the length of $\vec v$ is $1$ and the length of $(2\sqrt 2,1)$ is $3$ this means that $1\cdot 3 \cdot \cos \theta= 3$, where $\theta$ is the angle between $\vec v$ and $(2\sqrt 1,1)$.
But this can only be true if $\cos\theta=1$, or in other words $\theta=0$, so $\vec v$ is parallel to $(2\sqrt 2,1)$, and therefore
$$ x=\arctan\frac{1}{2\sqrt 2} \qquad\text{or, equivalently,}\qquad
x = \arcsin\frac13 $$
(plus any multiple of $2\pi$, of course).
A: $$\sin  x+2\sqrt { 2 } \cos { x } =3\\ 2\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } +2\sqrt { 2 } \left( \cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 }  }  }  \right) =3\sin ^{ 2 }{ \frac { x }{ 2 } +3\cos ^{ 2 }{ \frac { x }{ 2 }  }  }  }  } \\ \left( 3+2\sqrt { 2 }  \right) \sin ^{ 2 }{ \frac { x }{ 2 }  } -2\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 }  }  } +\left( 3-2\sqrt { 2 }  \right) \cos ^{ 2 }{ \frac { x }{ 2 }  } =0\\ \left( 3+2\sqrt { 2 }  \right) \tan ^{ 2 }{ \frac { x }{ 2 }  } -2\tan { \frac { x }{ 2 }  } +\left( 3-2\sqrt { 2 }  \right) =0\\ \tan { \frac { x }{ 2 } =\frac { 2\pm \sqrt { 3 }  }{ 2\left( 3+2\sqrt { 2 }  \right)  }  } \\ \\ \\ \\  $$
Can you take here?
A: 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
$$x=(\sin^{-1}\frac{c}{R})- \theta = (\sin^{-1}\frac{c}{R})- (\sin^{-1}\frac{b}{R}) = (\sin^{-1}\frac{3}{3})- (\sin^{-1}\frac{2\sqrt{2}}{3}) = 0.3398$$ 
A: 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:
$$
x=\arctan\dfrac{1}{2\sqrt2}+2k\pi,\quad\text{for $k\in\mathbf{Z}$}.
$$
A: I would emphasize that
$$ \sin (x + A) = \sin x \cos A + \cos x \sin A  $$
You have
$$ \sin x \; \; \frac{1}{3} + \cos x \; \;  \frac{\sqrt 8}{3} = 1   $$
We can take $A = \arctan \sqrt 8,$ so that $\cos A = 1/3$ and $\sin A = \sqrt 8 /  3. $ So, once again, you have
$$ \sin (x + \arctan \sqrt 8) = 1.  $$
On e of the answers is
$$ x = \frac{\pi}{2} - \arctan \sqrt 8 $$
A: Rewrite the equation as
$$3\Bigl(\frac13\sin x+\frac{2\sqrt2}3\cos x \Bigr)=3$$
Set $\varphi=\arctan \Bigl(-\dfrac 1{2\sqrt 2}\Bigr)=\arctan\Bigl(-\dfrac{\sqrt2}4\Bigr)$. This equation becomes
$$\cos(x-\varphi)=1\iff x-\varphi\equiv 0\mod 2\pi\iff x\equiv \arctan\Bigl(-\frac{\sqrt 2}4\Bigr)\mod 2\pi.$$
