Solve $x-1 \ge \sin(x)$ I got as far as $-1 = \sin(x)-x$.
I don't know what to do next. Pretty sure I am forgetting some simplification rule.
 A: There is no analytical solution to the problem. 
Numerical solution suggests $x\ge 1.9345632$
A: Let:
$$a_{n+1}=\sin(a_n)+1$$
$$b_{n+1}=\sin(b_n+1)$$
Then the point of equality $p$ is given by:
$$p=\lim_{n\to\infty}a_n=\bigg[\lim_{n\to\infty} b_n\bigg]+1$$
A: As already said in comments and answers, there is no analytical solution.
By inspection or graphing, you should notice that the first root is "close" to $\frac{2\pi} 3$. So, using Taylor expansion around this value
$$\sin(x)-x+1=\left(1+\frac{\sqrt{3}}{2}-\frac{2 \pi }{3}\right)-\frac{3}{2} \left(x-\frac{2 \pi
   }{3}\right)+O\left(\left(x-\frac{2 \pi }{3}\right)^2\right)$$ Ignoring the higher order terms, an approximation is
$$x=\frac{6+3 \sqrt{3}+2 \pi}{9} \approx 1.94215$$ Using on more term in the expansion will give a quadratic equation in $\left(x-\frac{2 \pi }{3}\right)$ and the result would be $\approx 1.93480$ which is quite close to the "exact" value given by Mohammad Riazi-Kermani.
A: Others have suggested there is no analytical solutions, however;
Let $x=\frac{5\pi }{2}$, then $\frac{5\pi }{2}-1\geq \sin(\frac{5\pi }{2})=1$. Just an example, you can go from here.
It seems there are infinite solutions.
A: It seems analytic solution is more subtle .so,by numerical methods ,You can try Newton-Rephson method ,


*

*F(X)=sinx-x+1=0.

*X2=x1-(f(x1)/f'(x1)). X1 is initial point.

*x1 =(a+b)/2, where f(a)*f(b)<0.

*as per required accuracy iteration can be done.

