What is the general solution for (sin(x))/(1-cos(x))=0 ?

The general solution for (sin(x))/(1-cos(x))=0 is x=pi+2pin

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

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

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

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

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Since the equation is undefined for:

2 πn

x = π + 2 πn

2 sec^{2} (x) - 3 + tan (x) = 0

sin (3 x + 87) - cos (7 x - 17) = 0

sin (20 x) = 0

sin (8 x) + sin (4 x) - 0.5 sin (6 x) = 0

4 sin^{2} (θ) + 4 cos (θ) = 0

What is the general solution for (sin(x))/(1-cos(x))=0 ?
The general solution for (sin(x))/(1-cos(x))=0 is x=pi+2pin