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

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

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(sin(x))/(2+cos(x)-1)=0

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

x = 2 πn

Degrees

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

Solution steps

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

π + 2 πn

x = 2 πn

y, arcsin (y) = ln (x)

sin (9 x + 6) = cos (3 x - 4)

csc (x) = \frac{1}{sec ( x )}

sin (x + 3 0^{\circ}) = cos (2 x)

tan (2 θ) = 0.55

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