What is the general solution for solvefor x,arcsin(x+y)=0 ?

The general solution for solvefor x,arcsin(x+y)=0 is x=-y

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solvefor x,arcsin(x+y)=0

solve for

x, arcsin (x + y) = 0

x = - y

Solution steps

arcsin (x + y) = 0

Apply trig inverse properties

arcsin (x + y) = 0

arcsin (x) = a \Rightarrow x = sin (a)

x + y = sin (0)

sin (0) = 0

sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0

x + y = 0

x + y = 0

Solve

x + y = 0 : x = - y

x + y = 0

Move

y

to the right side

x + y = 0

Subtract

y

from both sides

x + y - y = 0 - y

Simplify

x = - y

x = - y

x = - y

1 + cos (π x) = sin (π x)

sin (x - 1 5^{\circ}) = cos (2 0^{\circ} + x)

15 sin^{2} (x) - 17 sin (x) + 4 = 0

cos (x) = \frac{451.1}{497}

tan^{2} (x) = 2 sec (x)

What is the general solution for solvefor x,arcsin(x+y)=0 ?
The general solution for solvefor x,arcsin(x+y)=0 is x=-y