Applications of calculus We have the following formula for area
$$A = r^2(\sinθ\cosθ-\sqrt{3}\sin(θ)^2)$$
We then need to find what value θ will give maximum area, so we differentiate to get;
$$
\frac{\mathrm{d}A}{\mathrm{d}θ} = r^2((-\sinθ)^2+(\cosθ^2-2\sqrt{3}\sin(θ)^2\cosθ)
$$
but how do I simplify this to find the turning points and hence the maximum value of θ?
So far I have simplified to;
$$
1(1+2\sqrt{\cos\theta}) = (\tan\theta)^2
$$
but I'm not sure if this is the right way to go about it as I have no idea where to go from here.
Thanks in advance!
 A: I do not know whether you set up the problem correctly, so cannot guarantee that $r$ is constant, and that therefore we must maximize $\sin\theta\cos\theta-\sqrt{3}\sin^2\theta$.
An exact description of the actual problem would be useful. 
But let us maximize. The derivative of $\sin\theta\cos\theta$, by the Product Rule, is $(\sin\theta)(-\sin\theta)+(\cos\theta)(\cos\theta)$, which is $\cos^2\theta-\sin^2\theta$. 
The derivative of $\sin^2\theta$, by the Product Rule or Chain Rule, is $2\sin\theta\cos\theta$.
So our derivative is $\cos^2\theta-\sin^2\theta-2\sqrt{3}\sin\theta\cos\theta$.
Set this equal to $0$. We cannot have $\cos\theta=0$, so we can divide by $\cos^2\theta$, and change signs, obtaining $\tan^2\theta+2\sqrt{3}\tan\theta-1=0$. This is a quadratic equation in $\tan\theta$.  The Quadratic Formula works out very nicely here, as does completing the square. 
Remark: Because of uncertainty about the placement of brackets, I do not know whether you mean $\sqrt{3}\sin^2\theta$ or $\sqrt{3}\sin(\theta^2)$. If it is the latter, then the derivative of the last part would be $2\sqrt{3}\theta\cos(\theta^2)$, and we would obtain an equation that is hopeless to solve exactly. 
There is in the post a problem with the treatment of parentheses. That probably is a large part of the reason for the difficulties in computation.
A: Hint
Continuing in the same spirit as André Nicolas, assuming that $$A=r^2 \Big(\sin (t) \cos (t)-\sqrt{3} \sin ^2(t)    \Big)$$ deriving what is inside the brackets with respect to $t$ gives $$A'=r^2 \Big(-\sin ^2(t)+\cos ^2(t)-2 \sqrt{3} \sin (t) \cos (t)  \Big)$$ which can be rearranged as $$A'=r^2 \Big(\cos (2 t)-\sqrt{3} \sin (2 t)  \Big)$$ So, the maximum of A corresponds to the solution of $$\tan(2t)=\frac{1}{\sqrt{3}}$$ that is to say $$2t=\frac{\pi}{6}+2k\pi$$ At this point, in order to simplify further calculations, it looks better to use $\sin (t) \cos (t)=\frac{1}{2}\sin(2t)$ and $\sin ^2(t)=\frac{1}{2} (1-\cos (2 t))$ which allow to rewrite $$A=r^2 \Big(\frac{1}{2} \sin (2 t)+\frac{1}{2} \sqrt{3} \cos (2 t)-\frac{\sqrt{3}}{2}\Big)$$ what could have been done from the very beginning.
Further simplifications could be done but I let them to you.
I am sure that you can take from here.
A: A little Trigonometric manipulation at the start will ease of the calculation
$$\frac{2A}{r^2}=2\sin\theta\cos\theta-\sqrt3(2\sin^2\theta)=\sin2\theta-\sqrt3(1-\cos2\theta)$$
$$\implies\frac{2A}{r^2}=\sin2\theta+\sqrt3\cos2\theta-\sqrt3$$
$$\frac2{r^2}\frac{d(A)}{d\theta}=2\cos2\theta-2\sqrt3\sin\theta$$
I think the rest should not be too difficult to deal with
