What is the general solution for solvefor x,z=sin^2(sqrt(x)-2piy) ?

The general solution for solvefor x,z=sin^2(sqrt(x)-2piy) is

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solvefor x,z=sin^2(sqrt(x)-2piy)

solve for

x, z = sin^{2} (x - 2 π y)

x = 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (z) + 4 π^{2} n^{2} + arcsin^{2} (z) + 4 πn arcsin (z), x = 4 π^{2} y + 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (- z) + π^{2} + 4 π^{2} n + 4 π^{2} n^{2} + 2 π arcsin (- z) + arcsin^{2} (- z) + 4 πn arcsin (- z), x = 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (- z) + 4 π^{2} n^{2} + arcsin^{2} (- z) + 4 πn arcsin (- z), x = 4 π^{2} y + 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (z) + π^{2} + 4 π^{2} n + 4 π^{2} n^{2} + 2 π arcsin (z) + arcsin^{2} (z) + 4 πn arcsin (z)

Solution steps

z = sin^{2} (x - 2 π y)

Switch sides

sin^{2} (x - 2 π y) = z

Solve by substitution

sin (x - 2 π y) = z, sin (x - 2 π y) = - z

sin (x - 2 π y) = z : x = 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (z) + 4 π^{2} n^{2} + arcsin^{2} (z) + 4 πn arcsin (z), x = 4 π^{2} y + 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (- z) + π^{2} + 4 π^{2} n + 4 π^{2} n^{2} + 2 π arcsin (- z) + arcsin^{2} (- z) + 4 πn arcsin (- z)

sin (x - 2 π y) = - z : x = 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (- z) + 4 π^{2} n^{2} + arcsin^{2} (- z) + 4 πn arcsin (- z), x = 4 π^{2} y + 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (z) + π^{2} + 4 π^{2} n + 4 π^{2} n^{2} + 2 π arcsin (z) + arcsin^{2} (z) + 4 πn arcsin (z)

Combine all the solutions

x = 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (z) + 4 π^{2} n^{2} + arcsin^{2} (z) + 4 πn arcsin (z), x = 4 π^{2} y + 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (- z) + π^{2} + 4 π^{2} n + 4 π^{2} n^{2} + 2 π arcsin (- z) + arcsin^{2} (- z) + 4 πn arcsin (- z), x = 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (- z) + 4 π^{2} n^{2} + arcsin^{2} (- z) + 4 πn arcsin (- z), x = 4 π^{2} y + 4 π^{2} y^{2} + 8 π^{2} n y + 4 π y arcsin (z) + π^{2} + 4 π^{2} n + 4 π^{2} n^{2} + 2 π arcsin (z) + arcsin^{2} (z) + 4 πn arcsin (z)

sin (θ) = \frac{3}{3 5}

6 cos (θ) sin (θ) + cos (θ) = 0

cos (x) + cos (5 x) = 0

7 cos (t) + 5 sin (t) = 0

cos (X) = 0.4848

What is the general solution for solvefor x,z=sin^2(sqrt(x)-2piy) ?
The general solution for solvefor x,z=sin^2(sqrt(x)-2piy) is