Trying to find solutions to the equation of $$(1/x)+1=2\sinx+3$$ and I have to rearrange it into the form of $x=f(x)$ as I am using the fixed point method and iterations.
Struggling to find new arrangements and I have these rearrangements already
$$x=\sin^{-1}((1/2x)-1)$$
$$x=1/(2\sin x+2)$$
$$x=x(2\sin x)+3x-1$$
Thanks so much!
If we square both sides, we get
$$\frac1{x^2}+\frac2x+1=4\sin^2(x)+12\sin(x)+9$$
And from here, we can get many forms for $x=f(x)$,
Solving for $\frac1{x^2}$:
$$\implies\frac1{x^2}=4\sin^2(x)+12\sin(x)+8-\frac2x$$
$$\implies x=\frac{\pm1}{\sqrt{4\sin^2(x)+12\sin(x)+8-\frac2x}}\tag{use ONLY $+$ or $-$}$$
Solving for $\frac2x$
$$\implies\frac2x=4\sin^2(x)+12\sin(x)+8-\frac1{x^2}$$
$$\implies x=\frac1{2\sin^2(x)+6\sin(x)+4-\frac1{2x^2}}$$
And you can get many more from solving on the $\sin$ side.