What is the general solution for a=((1+sin^2(x)))/((1-sin^2(x))) ?

The general solution for a=((1+sin^2(x)))/((1-sin^2(x))) is

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a=((1+sin^2(x)))/((1-sin^2(x)))

a = \frac{( 1 + sin ^{2} ( x ) )}{( 1 - sin ^{2} ( x ) )}

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

Solution steps

a = \frac{( 1 + sin ^{2} ( x ) )}{( 1 - sin ^{2} ( x ) )}

Switch sides

\frac{1 + sin ^{2} ( x )}{1 - sin ^{2} ( x )} = a

Solve by substitution

sin (x) = \frac{- 1 + a}{1 + a}, sin (x) = - \frac{- 1 + a}{1 + a}; a \neq = - 1

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

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

Combine all the solutions

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

x, b \cdot f = sin^{3} (x)

2 sin^{2} (x) + sin^{3} (x) - 1 = 0

2 cos^{2} (x) = 3 cos (x) - 1

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

3 cos^{2} (x) - 10 cos (x) + 3 = 0

What is the general solution for a=((1+sin^2(x)))/((1-sin^2(x))) ?
The general solution for a=((1+sin^2(x)))/((1-sin^2(x))) is