What is the general solution for solvefor x,sin^2(x)=((1-cos(y)))/2 ?

The general solution for solvefor x,sin^2(x)=((1-cos(y)))/2 is

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solvefor x,sin^2(x)=((1-cos(y)))/2

solve for

x, sin^{2} (x) = \frac{( 1 - cos ( y ) )}{2}

x = arcsin (\frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn, x = arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (\frac{1 - cos ( y )}{2}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 1 - cos ( y ) )}{2}

Solve by substitution

sin^{2} (x) = \frac{1 - cos ( y )}{2}

Let:

sin (x) = u

u^{2} = \frac{1 - cos ( y )}{2}

u^{2} = \frac{1 - cos ( y )}{2} : u = \frac{1 - cos ( y )}{2}, u = - \frac{1 - cos ( y )}{2}

u^{2} = \frac{1 - cos ( y )}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1 - cos ( y )}{2}, u = - \frac{1 - cos ( y )}{2}

Substitute back

u = sin (x)

sin (x) = \frac{1 - cos ( y )}{2}, sin (x) = - \frac{1 - cos ( y )}{2}

sin (x) = \frac{1 - cos ( y )}{2}, sin (x) = - \frac{1 - cos ( y )}{2}

sin (x) = \frac{1 - cos ( y )}{2} : x = arcsin (\frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn

sin (x) = \frac{1 - cos ( y )}{2}

Apply trig inverse properties

sin (x) = \frac{1 - cos ( y )}{2}

General solutions for

sin (x) = \frac{1 - cos ( y )}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn

x = arcsin (\frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn

sin (x) = - \frac{1 - cos ( y )}{2} : x = arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (\frac{1 - cos ( y )}{2}) + 2 πn

sin (x) = - \frac{1 - cos ( y )}{2}

Apply trig inverse properties

sin (x) = - \frac{1 - cos ( y )}{2}

General solutions for

sin (x) = - \frac{1 - cos ( y )}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (\frac{1 - cos ( y )}{2}) + 2 πn

x = arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (\frac{1 - cos ( y )}{2}) + 2 πn

Combine all the solutions

x = arcsin (\frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn, x = arcsin (- \frac{1 - cos ( y )}{2}) + 2 πn, x = π + arcsin (\frac{1 - cos ( y )}{2}) + 2 πn

sin (x) + sin^{2} (x) + cos (x) + cos^{2} (x) = 0

4 cos^{2} (x) - 4 sin (x) - 1 = 0

cos^{2} (x) + 1 = - 2 cos (x)

sin^{3} (x) - 2 sin (x) = 0

2 cos^{3} (x) = cot^{3} (x)

What is the general solution for solvefor x,sin^2(x)=((1-cos(y)))/2 ?
The general solution for solvefor x,sin^2(x)=((1-cos(y)))/2 is