Solve $3\sin^2 x - \cos^2 x - 2 =0$ 
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$ 

My attempt - 
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x  + \sin^2 x = 3 $
$4\sin^2 x = 3 $ 
$ \sin x= \frac{\sqrt{3}}{2} $ 
I found that $x= 60,120 $ 
Why is the answer for this $60,120,240,300$ ? How do I find 240 and 300? 

The original definition of $\sin \theta$ of a right-angled triangle is just $\sin \theta =\dfrac {\text {opposite side}}{\text {hypotenuse}}$.  
Later, because of its in-adequacy in handling angles larger than $90^0$, the definition is expanded to $\sin \theta = \dfrac {y-ordinate}{\text {radius of the unit circle}} = \dfrac {y-ordinate}{1}$ for the point P = (x, y).

As seen in the figure, P' = (-x, y) also satisfies that definition. Therefore, for a single valued ($\dfrac {\sqrt 3}{2}$), there corresponds two solutions of $\theta$.  Another two for  ($- \dfrac {\sqrt 3}{2})$.
