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

The general solution for (sin(x)+cos(x))/(sin(x))=1-1/(tan(x)) is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

Subtract

1 - \frac{1}{tan ( x )}

from both sides

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

Simplify

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

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

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

tan (x) cos (x) + sin (x) = 0

Rewrite using trig identities

2 sin (x) = 0

Divide both sides by

2

sin (x) = 0

General solutions for

sin (x) = 0

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 2 πn, x = π + 2 πn

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution for x \in R

6 cos^{2} (x) - 5 sin (x) - 2 = 0

sec (θ) = - \frac{7}{4}

arctan (x) = \infty

r = a (1 - sin (x))

(cos (θ)) (cos (θ)) + 1 = (sin (θ)) (sin (θ))

What is the general solution for (sin(x)+cos(x))/(sin(x))=1-1/(tan(x)) ?
The general solution for (sin(x)+cos(x))/(sin(x))=1-1/(tan(x)) is No Solution for x\in\mathbb{R}