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

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

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(cot^2(x))/(1+sin(x))=(csc(x)-1)/(sin^2(x))

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

x = \frac{π}{2} + 2 πn

Solution steps

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

Subtract

\frac{csc ( x ) - 1}{sin ^{2} ( x )}

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

- 1 + csc (x) = 0 : x = \frac{π}{2} + 2 πn

- 1 - sin (x) + sin^{2} (x) + sin^{2} (x) csc (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

tan (x) = 2.3

2 sec^{2} (x) = 5 tan (x) + 5

tan^{2} (x) - 2 tan (x) = 3

x, y = arcsin (x y)

9 cos^{2} (θ) - 2 sin (θ) - 9 = 6 sin (θ) - 1

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