What is the general solution for solvefor x,z=arcsin^3(3x+2y) ?

The general solution for solvefor x,z=arcsin^3(3x+2y) is x=(sin(\sqrt[3]{z}))/3-(2y)/3

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solvefor x,z=arcsin^3(3x+2y)

solve for

x, z = arcsin^{3} (3 x + 2 y)

Solution steps

z = arcsin^{3} (3 x + 2 y)

Switch sides

arcsin^{3} (3 x + 2 y) = z

Solve by substitution

No Solution

Combine all the solutions

(3 sec (θ) - 23) (2 sin (θ) - 1) = 0, 2 π

7 cos (θ) - 3 = 3 sin^{2} (θ)

cos (\frac{π}{6} t) = cos (\frac{π}{6} (t - 3))

0 = arcsin (x)

cos (2 y) - sin (2 y) = 0

What is the general solution for solvefor x,z=arcsin^3(3x+2y) ?
The general solution for solvefor x,z=arcsin^3(3x+2y) is x=(sin(\sqrt[3]{z}))/3-(2y)/3