What is the inverse of 5+2sin((x-1)/(x+1)) ?

The inverse of 5+2sin((x-1)/(x+1)) is (arcsin((x-5)/2)+1)/(1-arcsin((x-5)/2))

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inverse of 5+2sin((x-1)/(x+1))

inverse

5 + 2 sin (\frac{x - 1}{x + 1})

\frac{arcsin ( \frac{x - 5}{2} ) + 1}{1 - arcsin ( \frac{x - 5}{2} )}

Solution steps

Solve for

sin (\frac{x - 1}{x + 1}), y = 5 + 2 sin (\frac{x - 1}{x + 1})

sin (\frac{x - 1}{x + 1}) = \frac{y - 5}{2}

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

\frac{x - 1}{x + 1} = arcsin (\frac{y - 5}{2})

Solve for

x, \frac{x - 1}{x + 1} = arcsin (\frac{y - 5}{2})

\frac{arcsin ( \frac{y - 5}{2} ) + 1}{1 - arcsin ( \frac{y - 5}{2} )}

Substitute

y = x

\frac{arcsin ( \frac{x - 5}{2} ) + 1}{1 - arcsin ( \frac{x - 5}{2} )}

What is the inverse of 5+2sin((x-1)/(x+1)) ?
The inverse of 5+2sin((x-1)/(x+1)) is (arcsin((x-5)/2)+1)/(1-arcsin((x-5)/2))