What is the general solution for sin(z)=0 ?

The general solution for sin(z)=0 is z=2pin,z=pi+2pin

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sin(z)=0

sin (z) = 0

z = 2 πn, z = π + 2 πn

Degrees

z = 0^{\circ} + 36 0^{\circ} n, z = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (z) = 0

General solutions for

sin (z) = 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}

z = 0 + 2 πn, z = π + 2 πn

z = 0 + 2 πn, z = π + 2 πn

Solve

z = 0 + 2 πn : z = 2 πn

z = 0 + 2 πn

0 + 2 πn = 2 πn

z = 2 πn

z = 2 πn, z = π + 2 πn

4 sec (x) + 4 tan (x) = 4

8 sin (2 x) sin (x) = 8 cos (x)

cos (x) = - \frac{3}{4}

2 cos^{2} (x) + cos (x) - 1 = 0, 0 \leq x \leq 2 π

arcsin (x) - arccos (- \frac{1}{2}) = \frac{π}{6}

What is the general solution for sin(z)=0 ?
The general solution for sin(z)=0 is z=2pin,z=pi+2pin