I'm having trouble finding the solution of the inverse of the function ${\rm f}\left(y\right) = \arcsin\left(\,3 - x^2\,\right)$
Isn't $\arcsin$ the inverse of $\sin$ ?. This is what I have now as inverse: $\sin\left(\,3 - x^2\,\right)$.
I'm having trouble finding the solution of the inverse of the function ${\rm f}\left(y\right) = \arcsin\left(\,3 - x^2\,\right)$
Isn't $\arcsin$ the inverse of $\sin$ ?. This is what I have now as inverse: $\sin\left(\,3 - x^2\,\right)$.
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$$y(x)=\arcsin(3-x^2)\iff x(y)=\sqrt{3-\sin(y)}$$.