What is the general solution for solvefor x,sin(x^2)+cos(y^3)=-2 ?

The general solution for solvefor x,sin(x^2)+cos(y^3)=-2 is

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solvefor x,sin(x^2)+cos(y^3)=-2

solve for

x, sin (x^{2}) + cos (y^{3}) = - 2

x = arcsin (- 2 - cos (y^{3})) + 2 πn, x = - arcsin (- 2 - cos (y^{3})) + 2 πn, x = π + arcsin (2 + cos (y^{3})) + 2 πn, x = - π + arcsin (2 + cos (y^{3})) + 2 πn

Solution steps

sin (x^{2}) + cos (y^{3}) = - 2

Move

cos (y^{3})

to the right side

sin (x^{2}) = - 2 - cos (y^{3})

Apply trig inverse properties

x^{2} = arcsin (- 2 - cos (y^{3})) + 2 πn, x^{2} = π + arcsin (2 + cos (y^{3})) + 2 πn

Solve

x^{2} = arcsin (- 2 - cos (y^{3})) + 2 πn : x = arcsin (- 2 - cos (y^{3})) + 2 πn, x = - arcsin (- 2 - cos (y^{3})) + 2 πn

Solve

x^{2} = π + arcsin (2 + cos (y^{3})) + 2 πn : x = π + arcsin (2 + cos (y^{3})) + 2 πn, x = - π + arcsin (2 + cos (y^{3})) + 2 πn

x = arcsin (- 2 - cos (y^{3})) + 2 πn, x = - arcsin (- 2 - cos (y^{3})) + 2 πn, x = π + arcsin (2 + cos (y^{3})) + 2 πn, x = - π + arcsin (2 + cos (y^{3})) + 2 πn

8 sin (12 0^{\circ} - x) = 9 sin (x)

cos (θ) = 0.625

x, sin (x) = 3 cos (x)

x, 1.12 = 0.04 sin (x)

sin (2 x) = - 0.8

What is the general solution for solvefor x,sin(x^2)+cos(y^3)=-2 ?
The general solution for solvefor x,sin(x^2)+cos(y^3)=-2 is