Find all the angles $v$ between $-\pi$ and $\pi$ Find all the angles $v$ between $-\pi$ and $\pi$ such that
$$-\sin(v)+ \sqrt3 \cos(v) = \sqrt2$$
The answer has to be in the form of: $\pi/2$ (it must include $\pi$)
I have tried squaring but I get nowhere.
 A: It's sometimes a good idea to turn a sum of sines and cosines into a single trigonometric term when solving equations of the form $a \sin x + b\cos x = c$, in this case: $$\sqrt{3}\cos v - \sin v \equiv 2\cos \left(v + \frac{\pi}{6}\right)$$
So, if you set $x = v + \frac{\pi}{6}$ you need only solve $\cos x = \frac{1}{\sqrt{2}}$ which you can then do, I'm sure.
A: Divide the equation by $\;2\;$ :
$$\frac1{\sqrt2}=-\frac12\sin x+\frac{\sqrt3}2\cos x=\sin\frac\pi3\cos x-\cos\frac\pi3\sin x=\sin\left(\frac\pi3-x\right)\implies$$
$$\frac\pi3-x=\begin{cases}\cfrac\pi4\\{}\\\cfrac{3\pi}4\end{cases}\implies \ldots$$
Observe this is similar to the other answer but, perhaps, a little, very little, easier to understand.
A: As noted by another user, it's useful to combine a sum of sines and cosines into a single function.
$$
\sqrt{3}\cos(v) - \sin(v) = \sqrt{2}
$$
In order to turn the LHS into a single function we have a bit of a choice here we can make the single function a sine or a cosine, and we can also choose if we want an addition or subtraction in the angle.
For this I'll show how to do it with a subtraction in the angle and a cosine function.
Recall from the angle difference formula for cosine
$$
\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)
$$
Now we set the equality
$$
\begin{align}
\sqrt{3}\cos(v) - \sin(v) &= A\cos(v -\theta)\\
\sqrt{3}\cos(v) - \sin(v) &= A \left [ \cos(v)\cos(\theta) + \sin(v)\sin(\theta) \right ]\\
\sqrt{3}\cos(v) - \sin(v) &= A\cos(\theta)\cos(v) + A\sin(\theta)\sin(v)
\end{align}
$$
From here we equate the coefficients
$$
\begin{align}
A\cos(\theta) &= \sqrt{3}\\
A\sin(\theta) &= -1
\end{align}
$$
Solving for $\theta$ gives
$$
\tan(\theta) = \frac{-1}{\sqrt{3}} \implies \theta = \frac{-\pi}{6}
$$
Solving for $A$ gives
$$
A\cos \left ( \frac{-\pi}{6} \right ) = \sqrt{3} \implies A\frac{\sqrt{3}}{2} = \sqrt{3} \implies A = 2
$$
Therefore, 
$$
\sqrt{3}\cos(v) - \sin(v) = 2\cos(v +\frac{\pi}{6})
$$
Just make this substitution, and I'm sure you can do the rest.
