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

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

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

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

No Solution for x \in R

Solution steps

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

Subtract

tan (x)

from both sides

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

Simplify

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

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

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

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

Factor

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

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

Solving each part separately

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

tan (x) = 0 : x = πn

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

Combine all the solutions

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

Since the equation is undefined for:

πn, \frac{π}{2} + 2 πn, 2 πn + π

No Solution for x \in R

sec (x + 4 5^{\circ}) = 1

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

- 2 sin (θ) = 0

sec^{3} (x) = - 1

\frac{1}{cot ^{2} ( x )} + 3 tan (x) = 0

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