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

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

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inverse of f(x)=1+sin^2(x-e/2)

inverse

f (x) = 1 + sin^{2} (x - \frac{e}{2})

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

Solution steps

Solve for

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

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

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

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

Solve for

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

Solve for

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

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

Substitute

y = x

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

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