x+arcsin(y)=xySubtract&nbsp;xy&nbsp;from both sides of the equation.x+arcsin(y)-xy=0Subtract&nbsp;arcsin(y)&nbsp;from both sides of the equation.x-xy=-arcsin(y)Factor&nbsp;x&nbsp;out of&nbsp;x-xy.Factor&nbsp;x&nbsp;out of&nbsp;x1.x⋅1-xy=-arcsin(y)Factor&nbsp;x&nbsp;out of&nbsp;-xy.x⋅1+x(-1y)=-arcsin(y)Factor&nbsp;x&nbsp;out of&nbsp;x⋅1+x(-1y).x(1-1y)=-arcsin(y)Rewrite&nbsp;-1y&nbsp;as&nbsp;-y.x(1-y)=-arcsin(y)Divide each term in&nbsp;x(1-y)=-arcsin(y)&nbsp;by&nbsp;1-y&nbsp;and simplify.Divide each term in&nbsp;x(1-y)=-arcsin(y)&nbsp;by&nbsp;1-y.x(1-y)1-y=-arcsin(y)1-ySimplify the left side.x=-arcsin(y)1-ySimplify the right side.x=-arcsin(y)1-y

We have an answer to "x+arcsiny=xy"

High School

maheynbaloch

Answered question

2022-04-10

x+arcsiny=xy

Answer & Explanation

nick1337

Expert2022-06-08Added 777 answers

$x + \arcsin (y) = x y$

Subtract $x y$ from both sides of the equation.

$x + \arcsin (y) - x y = 0$

Subtract $\arcsin (y)$ from both sides of the equation.

$x - x y = - \arcsin (y)$

Factor $x$ out of $x - x y$ .

Factor $x$ out of $x^{1}$ .

$x \cdot 1 - x y = - \arcsin (y)$

Factor $x$ out of $- x y$ .

$x \cdot 1 + x (- 1 y) = - \arcsin (y)$

Factor $x$ out of $x \cdot 1 + x (- 1 y)$ .

$x (1 - 1 y) = - \arcsin (y)$

Rewrite $- 1 y$ as $- y$ .

$x (1 - y) = - \arcsin (y)$

Divide each term in $x (1 - y) = - \arcsin (y)$ by $1 - y$ and simplify.

Divide each term in $x (1 - y) = - \arcsin (y)$ by $1 - y$ .

$\frac{x (1 - y)}{1 - y} = \frac{- \arcsin (y)}{1 - y}$

Simplify the left side.

$x = \frac{- \arcsin (y)}{1 - y}$

Simplify the right side.

$x = - \frac{\arcsin (y)}{1 - y}$

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