Solve the trigonometric series equation $1 + \sin x + \sin^2 x + \sin^3 x + \cdots = 4 + 2\sqrt 3$

Question

$$1 + \sin x + \sin^2 x + \sin^3 x + \cdots = 4 + 2\sqrt 3$$ Find $x$.

What I first did was I applied addition of geometric progression condition that I just copied off a book.

That got me something like $$\frac{1 - \sin^{\infty} x}{1 - \sin x} = 4 - 2 \sqrt 3,$$ and I got stuck.

Please tell me any step I've taken wrongly, or something new to help me solve the problem. I've recently started studying for IIT-JEE, and am a beginner who just stumbled upon this question. Please help me out.

Answer 1

$\forall x \in \mathbb R |\sin x|\le1$ and if $|u|<1$ then $1+u+u^2+u^3+...=\frac1{1-u}$

$$\frac1{1-\sin x}=4+2\sqrt3$$

$$\sin x=\frac{\sqrt3}{2}$$ $$x=(-1)^k\frac{\pi}{3}+\pi k, k \in \mathbb Z$$

Answer 2

Continuing after reading the comments,

You are to solve:$${1\over 1-\sin(x)}=4+2\sqrt 3$$

Now $1-\sin(x)$ can be written as $(\sin(x/2)-\cos(x/2))^2$ So proceed from here.

Also check that the right side is $(1+\sqrt 3)^2$