What is the inverse of f(x)=arcsin(1/(2x-1)) ?

The inverse of f(x)=arcsin(1/(2x-1)) is (1+sin(x))/(2sin(x))

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inverse of f(x)=arcsin(1/(2x-1))

inverse

f (x) = arcsin (\frac{1}{2 x - 1})

\frac{1 + sin ( x )}{2 sin ( x )}

Solution steps

arcsin (a) = b \Rightarrow a = sin (b)

\frac{1}{2 x - 1} = sin (y)

Solve for

x, \frac{1}{2 x - 1} = sin (y)

\frac{1 + sin ( y )}{2 sin ( y )}

Substitute

y = x

\frac{1 + sin ( x )}{2 sin ( x )}

f (x) = 8 x + 10

y = - 7 x - 5

\frac{2 s + 2}{s ^{2} + 2 s + 5}

f (x) = 3 + x^{2}

+ 15 x^{3} - 14

What is the inverse of f(x)=arcsin(1/(2x-1)) ?
The inverse of f(x)=arcsin(1/(2x-1)) is (1+sin(x))/(2sin(x))