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

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

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(cos(x))/(1+sin(x))=1-(sin(x))/(cos(x))

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

x = 2 πn

Solution steps

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

Subtract

1 - \frac{sin ( x )}{cos ( x )}

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

1 - cos (x) + sin (x) - cos (x) sin (x) = 0

Factor

1 - cos (x) + sin (x) - cos (x) sin (x) : - (cos (x) - 1) (sin (x) + 1)

- (cos (x) - 1) (sin (x) + 1) = 0

Solving each part separately

cos (x) - 1 = 0 or sin (x) + 1 = 0

cos (x) - 1 = 0 : x = 2 πn

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

Combine all the solutions

x = 2 πn, x = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

x = 2 πn

y, 2 cos (5 x + y) = 0

2 cos^{2} (x) + 15 sin (x) - 15 = 0

cos (2 x) = sin (x), - 2 π \leq x \leq 2 π

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

cos (x) = \frac{8}{11}

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