What is the general solution for solvefor x,arcsin(y)-arccos(x)=1 ?

The general solution for solvefor x,arcsin(y)-arccos(x)=1 is x=cos(-1+arcsin(y))

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solvefor x,arcsin(y)-arccos(x)=1

solve for

x, arcsin (y) - arccos (x) = 1

x = cos (- 1 + arcsin (y))

Solution steps

arcsin (y) - arccos (x) = 1

Move

arcsin (y)

to the right side

arcsin (y) - arccos (x) = 1

Subtract

arcsin (y)

from both sides

arcsin (y) - arccos (x) - arcsin (y) = 1 - arcsin (y)

Simplify

- arccos (x) = 1 - arcsin (y)

- arccos (x) = 1 - arcsin (y)

Divide both sides by

- 1

- arccos (x) = 1 - arcsin (y)

Divide both sides by

- 1

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

Simplify

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

Simplify

\frac{- arccos ( x )}{- 1} : arccos (x)

\frac{- arccos ( x )}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{arccos ( x )}{1}

Apply rule

\frac{a}{1} = a

= arccos (x)

Simplify

\frac{1}{- 1} - \frac{arcsin ( y )}{- 1} : - 1 + arcsin (y)

\frac{1}{- 1} - \frac{arcsin ( y )}{- 1}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

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

Apply rule

\frac{a}{1} = a

= - (- arcsin (y) + 1)

Distribute parentheses

= - 1 - (- arcsin (y))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + arcsin (y)

arccos (x) = - 1 + arcsin (y)

arccos (x) = - 1 + arcsin (y)

arccos (x) = - 1 + arcsin (y)

Apply trig inverse properties

arccos (x) = - 1 + arcsin (y)

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

x = cos (- 1 + arcsin (y))

x = cos (- 1 + arcsin (y))

- 6 = - 2 + 8 sin (θ + \frac{π}{4})

0 = 12 cos (\frac{π}{3} x) + 8

0 = sin (t) + 2 sin (2 t)

sin (x) = - 0.9

1 = tan (\frac{π}{4} + c)

What is the general solution for solvefor x,arcsin(y)-arccos(x)=1 ?
The general solution for solvefor x,arcsin(y)-arccos(x)=1 is x=cos(-1+arcsin(y))