How find this there exsit $m\in N^{+}$,such $\sin{x}+a_{m}\cdot \sqrt{3-2\cos{x}-2\sin{x}}=1,0
let $$a_{1}=1,a_{n+1}=\ln{(1+a_{n})}-\dfrac{1}{2}a_{n}$$
prove or disprove 

there exsit $m\in N^{+}$,such $$\sin{x}+a_{m}\cdot \sqrt{3-2\cos{x}-2\sin{x}}=1,0<x<2\pi$$ have no solution.

my idea: $$3-2\sin{x}-2\cos{x}=(\sin{x}-1)^2+(\cos{x}-1)^2$$
so
$$a_{m}=\dfrac{1-\sin{x}}{\sqrt{(1-\sin{x})^2+(1-\cos{x})^2}}=f(x)$$
so I think we must find the $f(x)$ range?
Thank you
 A: There exists a solution for every $m\in\mathbb{N}^+$.
Note that $$f(x)=\dfrac{1-\sin{x}}{\sqrt{(1-\sin{x})^2+(1-\cos{x})^2}}$$ is a continuous function on the interval $[0,\frac{\pi}{2}]$, since the numerator is continuous everywhere and the denominator is continuous everywhere and always nonzero (as $\sin x=\cos x=0$ is not possible).
However, $f(0)=1$ and $f(\frac{\pi}{2})=0$, so by the intermediate value theorem for any $a\in[0,1]$ there exists some $y\in[0,\frac{\pi}{2}]$ such that $f(y)=a$.
To complete the proof, we show inductively that $a_m\in[0,1]$. 
Our base case is easy: $a_1=1\in[0,1]$. Now assume $a_k\in[0,1]$. Let $g(x)=\ln(1+x)-\frac{1}{2}x$. Then $g''(x)=\frac{-1}{(1+x)^2}$ which is negative on $x\in[0,1]$. Hence $g(x)$ is concave down on this domain, so noting that $g(0)=0\geq0$ and $g(1)=\ln 2-\frac{1}{2}\geq0$ gives us that $g(x)$ is positive for any $x\in[0,1]$. This combined with the fact that $g(x)\leq\ln(x+1)<\ln e=1$ on this domain gives us that $0\leq g(x)\leq 1$ when $x\in[0,1]$, so $0\leq g(a_k)=a_{k+1}\leq1$, and our inductive proof is complete.
Q.E.D.
