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

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

Divide both sides by

- 1

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

Apply trig inverse properties

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))