Real Solutions to the Equation Can you please find all real solutions to the equation
$$2 + \sin (\theta )+\frac{(\sin(\theta))^2}{2}+ \cdots = 2$$
 A: I take a guess this is equivalent to
$$
\sum_{n\geq 1}\frac{\sin^n\theta}{n}=0.
$$
For another possible interpretation, see Maesumi's answer.
1)For $|\sin\theta|<1$ we get, using the Taylor expansion of $\ln(1-x)$,
$$
-\ln(1-\sin\theta)=0 \quad\Rightarrow \quad\sin\theta=0.
$$
So the solutions are $\pi\mathbb{Z}$.
2)For $\sin\theta=1$, the series diverges, the equation is not defined.
3)For $\sin\theta=-1$, we get $-\ln 2=0$, no solution.
You solution set is 
$$
\pi\mathbb{Z}.
$$
A: The geometric series adds to ${2 \over {1-\sin(\theta)/2}}$. If you set the sum equal to $2$ and simplify it results in $\sin(\theta)=0$, i.e $\theta=n\pi$.
A geometric series has the following look $a+ax+ax^2+ax^3+\cdots$. For $|x|<1$ the series adds up to $a \over {1-x}$. Here your $a=2$ and your $x=\sin\theta /2$.
If you provide too few terms of a series doubts grow as to what it might be!
A: This is an infinite G.P
Since the common ratio is  $\frac{(\sin(\theta))}{2}<1$
${2 \over {1-\sin(\theta)/2}}=2$
${{1-\sin(\theta)/2}}=1$
$\sin(\theta)/2=0$
$\sin(\theta)=0$
$\theta=n\pi$.
