Number of solutions of $3\sin^2 x+\cos^2 x+\sqrt{3} \sin x+\cos x+1=\sqrt{3}\sin x\cos x$ Find Number of solutions of 

$$3\sin^2 x+\cos^2 x+\sqrt{3} \sin x+\cos x+1=\sqrt{3}\sin x\cos x$$ in $\left[0 \:\: 10\pi\right]$

My Try:
The given equation is
$$2+2\sin^2 x+2\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\sin (2x)$$ $\implies$
$$3+2\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\sin (2x)+\cos (2x)$$
Any idea here?
 A: We can rewrite the equation as $\left(\sqrt 3 \sin x - \cos x \right)^2+ \left(1+ \sqrt 3 \sin x \right) \left(1+\cos x\right) = 0$
Let $a=1+ \sqrt 3 \sin x$ and $b=1+\cos x$
Then the equation is $(a-b)^2+ab = 0 \Rightarrow a^2-ab+b^2 =0$
But $a^2 -ab+b^2 = 0 \iff a=0,b=0 $ which is not possible.
Hence the equation has no solutions.
A: Expand out the difference (left side - right side) as 
$$ F(x) = 4-2\,  \cos^2 \left( x \right)  +\sqrt {3}\sin \left( 
x \right) +\cos \left( x \right) -\sqrt {3}\sin \left( x \right) \cos
 \left( x \right) 
$$
If $\sin(x) = s$, $\cos(x) = \pm \sqrt{1-s^2}$, and
$$ F(x) = 2 + 2 s^2 + \sqrt{3} s \pm (1-\sqrt{3} s) \sqrt{1-s^2}$$
In order for that to be $0$, we'd need
$$ \sqrt{1-s^2} =  \mp \frac{2 s^2 + \sqrt{3} s + 2}{1-\sqrt{3} s} $$
and thus $$1 - s^2 = \dfrac{(2s^2 + \sqrt{3} s + 2)^2}{(1-\sqrt{3} s)^2}$$
so that
$$ 7\,{s}^{4}+2\,\sqrt {3}{s}^{3}+6\,\sqrt {3}s+9\,{s}^{2}+3 = 0$$
This quartic has positive discriminant ($209088$), so its roots are either all real or all non-real.  In this case they turn out to be all non-real.
A: Let $c = \cos(x)$ and $s = \sin(x)$, solve for $c$:
$$3s^2 + c^2 + \sqrt 3 s + c + 1 = \sqrt 3 cs$$
$$c^2 + (1 - \sqrt 3 s)c + 3s^2 + \sqrt 3 s + 1 = 0$$
$$c = \frac{\sqrt 3 s - 1 \pm \sqrt{ (1 - \sqrt 3 s)^2 - 4(3s^2 + \sqrt 3 s + 1)}}{2}$$
Normally here you'd check the range on the discriminate, but the discriminate is a perfect square, so:
$$c = \frac{\sqrt 3 s - 1 \pm \sqrt{ -9s^2 - 6\sqrt 3 s - 3}}{2}$$
$$c = \frac{\sqrt 3 s - 1 \pm i\sqrt{(3s + \sqrt 3)^2}}{2}$$
$$c = \frac{\sqrt 3 s - 1 \pm i|3s + \sqrt 3|}{2}$$
So if $s$ is real then $c$ is complex, so no solutions.
