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The result holds true for any number, not necessarily $2016$ . Below, $2016$ is taken to be $n$ .

First evaluate the bracket.

$\displaystyle\sum_{i=1}^{n}\sin^{i}(x)\cos^{2}(x)$

$=\displaystyle\sum_{i=1}^{n}\sin^{i}(x)(1-\sin^{2}(x))$

$=\displaystyle\sum_{i=1}^{n}\sin^{i}(x)-\sin^{i+2}(x)$

$=(\sin(x)-\sin^{3}(x))+(\sin^{2}-\sin^{4}(x))+(\sin^{3}(x)-\sin^{5}(x))+\cdots$

$+(\sin^{n-2}(x)-\sin^{n}(x))+(\sin^{n-1}(x)-\sin^{n+1}(x))+(\sin^{n}(x)-\sin^{n+2}(x))$

$=\sin(x)+\sin^{2}(x)-\sin^{n+1}(x)-\sin^{n+2}(x)$

Now, the part before the bracket, when simplified, comes to be $\sin^{n+1}(x)+\sin^{n+2}(x)$ .

Hence we have

$\sin(x)+\sin^{2}(x)=\frac{3}{4}$

$4\sin^{2}(x)+4\sin(x)-3=0$

$4\sin^{2}(x)+6\sin(x)-2\sin(x)-3=0$

$2\sin(x)(2\sin(x)+3)-(2\sin(x)+3)=0$

$(2\sin(x)+3)(2\sin(x)-1)=0 \implies \sin(x)=-\frac{3}{2}~\text{or}~\sin(x)=\frac{1}{2}$

But as $\sin(x)\in[-1,1]$ , we can say that $\sin(x)=\frac{1}{2}$

$\therefore x=30^{\circ}$