What is the general solution for solvefor y,x=tan^2(y) ?

The general solution for solvefor y,x=tan^2(y) is y=arctan(sqrt(x))+pin,y=arctan(-sqrt(x))+pin

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solvefor y,x=tan^2(y)

solve for

y, x = tan^{2} (y)

y = arctan (x) + πn, y = arctan (- x) + πn

Solution steps

x = tan^{2} (y)

Switch sides

tan^{2} (y) = x

Solve by substitution

tan^{2} (y) = x

Let:

tan (y) = u

u^{2} = x

u^{2} = x : u = x, u = - x

u^{2} = x

For

x^{2} = f (a)

the solutions are

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

u = x, u = - x

Substitute back

u = tan (y)

tan (y) = x, tan (y) = - x

tan (y) = x, tan (y) = - x

tan (y) = x : y = arctan (x) + πn

tan (y) = x

Apply trig inverse properties

tan (y) = x

General solutions for

tan (y) = x

tan (x) = a \Rightarrow x = arctan (a) + πn

y = arctan (x) + πn

y = arctan (x) + πn

tan (y) = - x : y = arctan (- x) + πn

tan (y) = - x

Apply trig inverse properties

tan (y) = - x

General solutions for

tan (y) = - x

tan (x) = a \Rightarrow x = arctan (a) + πn

y = arctan (- x) + πn

y = arctan (- x) + πn

Combine all the solutions

y = arctan (x) + πn, y = arctan (- x) + πn

tan (θ) = \frac{7}{3}

sin (4 θ) - sin (2 θ) = 0

tan (x) sin (x) - tan (x) = 0

cos^{2} (θ) - 7 cos (θ) + 12 = 0

cos^{2} (x) = cos^{2} (\frac{x}{2})

What is the general solution for solvefor y,x=tan^2(y) ?
The general solution for solvefor y,x=tan^2(y) is y=arctan(sqrt(x))+pin,y=arctan(-sqrt(x))+pin