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**flyingsquirrel** Hi

$\displaystyle g\circ f(x)=\frac{1}{2\sin x -1}$ This is a fraction hence it exists iff the denominator does not equal 0 : $\displaystyle 2\sin x -1\neq 0 \Longleftrightarrow \sin x \neq \frac{1}{2} \Longleftrightarrow x \neq ?$

To find the range, start from $\displaystyle -1\leq \sin x \leq 1$ then multiply by 2 : $\displaystyle -2\leq 2\sin x \leq 2$ then add 1...

$\displaystyle g$ is one-to-one iff $\displaystyle g'(x)\neq0$ for any $\displaystyle x$ belonging to the domain of $\displaystyle g$. To find $\displaystyle g^{-1}$, you can write $\displaystyle y=\frac{1}{2x-1}$ and solve this for $\displaystyle x$ : it'll give $\displaystyle x=g^{-1}(y)$.