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

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

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

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

x = 2 πn, x = π + 2 πn

Solution steps

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

Subtract

\frac{1}{cos ( x )}

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = 1 : x = 2 πn

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

sin (θ) = \frac{4}{5}, 0 \leq θ \leq \frac{π}{2}

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

2 - cos^{2} (x) = 4 sin^{2} (\frac{1}{2} x)

\frac{( tan ( x ) + 2 sin ( x ) )}{( tan ( x ) - 2 sin ( x ) )} = 3

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

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