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

The general solution for (sin^2(x))/(1+cos(x))=1 is x= pi/2+2pin,x=(3pi)/2+2pin

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(sin^2(x))/(1+cos(x))=1

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

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

Solution steps

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

Subtract

1

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

- cos (x) - cos^{2} (x) = 0

Solve by substitution

cos (x) = - 1, cos (x) = 0

cos (x) = - 1 : x = π + 2 πn

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

π + 2 πn

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

1 + sin (x) = cos (x)

cos (2 x) = \frac{2}{π}

4 tan^{2} (x) + 8 sec (x) = 1

cos^{2} (θ) + sin (θ) = 2

3 tan^{2} (θ) + tan (θ) = 0

What is the general solution for (sin^2(x))/(1+cos(x))=1 ?
The general solution for (sin^2(x))/(1+cos(x))=1 is x= pi/2+2pin,x=(3pi)/2+2pin