What is the general solution for solvefor x,y= 1/pi arctan(x/s)+1/2 ?

The general solution for solvefor x,y= 1/pi arctan(x/s)+1/2 is x=tan((2piy-pi)/2)s

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solvefor x,y= 1/pi arctan(x/s)+1/2

solve for

x, y = \frac{1}{π} arctan (\frac{x}{s}) + \frac{1}{2}

x = tan (\frac{2 π y - π}{2}) s

Solution steps

y = \frac{1}{π} arctan (\frac{x}{s}) + \frac{1}{2}

Switch sides

\frac{1}{π} arctan (\frac{x}{s}) + \frac{1}{2} = y

Solve by substitution

arctan (\frac{x}{s}) = π y - \frac{π}{2}

Subtract

y

from both sides

\frac{1}{π} arctan (\frac{x}{s}) + \frac{1}{2} - y = 0

Simplify

\frac{1}{π} arctan (\frac{x}{s}) + \frac{1}{2} - y : \frac{2 arctan ( \frac{x}{s} ) + π - 2 π y}{2 π}

\frac{2 arctan ( \frac{x}{s} ) + π - 2 π y}{2 π} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 arctan (\frac{x}{s}) + π - 2 π y = 0

Solve by substitution

arctan (\frac{x}{s}) = \frac{2 π y - π}{2}

Let:

u = \frac{x}{s}

2 arctan (u) + π - 2 π y = 0

Move

2 π y

to the right side

2 arctan (u) + π = 2 π y

Move

π

to the right side

2 arctan (u) = 2 π y - π

Divide both sides by

2

arctan (u) = \frac{2 π y - π}{2}

Apply trig inverse properties

u = tan (\frac{2 π y - π}{2})

Substitute back

u = \frac{x}{s}

\frac{x}{s} = tan (\frac{2 π y - π}{2}) : x = tan (\frac{2 π y - π}{2}) s; s \neq = 0

x = tan (\frac{2 π y - π}{2}) s

tan^{2} (θ) + 8 tan (θ) + 6 = 0

3 - sin (θ) = 5 sin (θ)

sin (x) + sin (2 x) = 0, 0 \leq x \leq 2 π

8 sin (2 x) = 16 cos (x)

3 arccos (x) = π

What is the general solution for solvefor x,y= 1/pi arctan(x/s)+1/2 ?
The general solution for solvefor x,y= 1/pi arctan(x/s)+1/2 is x=tan((2piy-pi)/2)s