What is the inverse of y=cos(x^2) ?

The inverse of y=cos(x^2) is sqrt(arccos(x)),-sqrt(arccos(x))

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inverse of y=cos(x^2)

inverse

y = cos (x^{2})

arccos (x), - arccos (x)

Solution steps

cos (a) = b \Rightarrow a = arccos (b)

x^{2} = arccos (y)

Solve for

x, x^{2} = arccos (y)

arccos (y), - arccos (y)

Substitute

y = x

arccos (x), - arccos (x)

f (x) = \frac{2}{1 + e ^{[x]}}

\frac{3 s - 14}{s ^{2} - 4 s + 8}

f (x) = - \frac{x - 17}{12}

f (x) = \frac{4 x + 1}{5 x - 2}

- \frac{3}{4} x + 12

What is the inverse of y=cos(x^2) ?
The inverse of y=cos(x^2) is sqrt(arccos(x)),-sqrt(arccos(x))