What is the general solution for solvefor x,sin(x-40)=cos(y-20) ?

The general solution for solvefor x,sin(x-40)=cos(y-20) is x=-y+360n+150,x=y+180+360n-70

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solvefor x,sin(x-40)=cos(y-20)

solve for

x, sin (x - 4 0^{\circ}) = cos (y - 2 0^{\circ})

x = - y + 36 0^{\circ} n + 15 0^{\circ}, x = y + 18 0^{\circ} + 36 0^{\circ} n - 7 0^{\circ}

Solution steps

sin (x - 4 0^{\circ}) = cos (y - 2 0^{\circ})

Rewrite using trig identities

sin (x - \frac{2 π}{9}) = sin (\frac{π}{2} - (y - \frac{π}{9}))

Apply trig inverse properties

x - \frac{2 π}{9} = \frac{π}{2} - (y - \frac{π}{9}) + 2 πn, x - \frac{2 π}{9} = π - (\frac{π}{2} - (y - \frac{π}{9})) + 2 πn

x - 4 0^{\circ} = 9 0^{\circ} - (y - 2 0^{\circ}) + 36 0^{\circ} n : x = - y + 36 0^{\circ} n + 15 0^{\circ}

x - 4 0^{\circ} = 18 0^{\circ} - (9 0^{\circ} - (y - 2 0^{\circ})) + 36 0^{\circ} n : x = y + 18 0^{\circ} + 36 0^{\circ} n - 7 0^{\circ}

x = - y + 36 0^{\circ} n + 15 0^{\circ}, x = y + 18 0^{\circ} + 36 0^{\circ} n - 7 0^{\circ}

0 = - sin (2 x)

1 - sin^{2} (θ) = 0

x, z = y sin (x y)

2 sin^{2} (x) + csc^{2} (x) = 3

t, tan (t) = 2

What is the general solution for solvefor x,sin(x-40)=cos(y-20) ?
The general solution for solvefor x,sin(x-40)=cos(y-20) is x=-y+360n+150,x=y+180+360n-70