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

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

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

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

x = \frac{π}{4} + πn

Solution steps

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

Subtract

- tan (x)

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Factor

- sin^{2} (x) + sin^{2} (x) tan (x) : sin^{2} (x) (tan (x) - 1)

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

Solving each part separately

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

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

tan (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^{6} (x) = - cos^{2} (x)

2 sin^{3} (x) - 5 sin^{2} (x) + 2 sin (x) = 0

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

\frac{sin ( x ) - ( 2 )}{2} = 0

cos (2 x) = 5 - 6 cos^{2} (x)

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