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

The general solution for (sin^2(x))/(1-cos^2(x))=cot(x) is x= pi/4+pin

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(sin^2(x))/(1-cos^2(x))=cot(x)

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

x = \frac{π}{4} + πn

Solution steps

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

Subtract

cot (x)

from both sides

\frac{sin ^{2} ( x )}{1 - cos ^{2} ( x )} - cot (x) = 0

Simplify

\frac{sin ^{2} ( x )}{1 - cos ^{2} ( x )} - cot (x) : \frac{sin ^{2} ( x ) - cot ( x ) ( 1 - cos ^{2} ( x ) )}{1 - cos ^{2} ( x )}

\frac{sin ^{2} ( x ) - cot ( x ) ( 1 - cos ^{2} ( x ) )}{1 - cos ^{2} ( x )} = 0

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

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

Rewrite using trig identities

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

Factor

sin^{2} (x) - cot (x) sin^{2} (x) : - sin^{2} (x) (cot (x) - 1)

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

Solving each part separately

sin^{2} (x) = 0 or cot (x) - 1 = 0

sin^{2} (x) = 0 : x = 2 πn, x = π + 2 πn

cot (x) - 1 = 0 : x = \frac{π}{4} + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

Since the equation is undefined for:

2 πn, π + 2 πn

x = \frac{π}{4} + πn

cos^{2} (x) + cos^{3} (x) + cos^{4} (x) + cos^{5} (x) = 0

sin^{2} (x) - 4 sin (x) = - 3

1 + 7 sinh (x) = 4 cosh^{2} (x)

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

cos^{2} (x) + 6 cos (x) + 5 = 0

What is the general solution for (sin^2(x))/(1-cos^2(x))=cot(x) ?
The general solution for (sin^2(x))/(1-cos^2(x))=cot(x) is x= pi/4+pin