What is the general solution for sin(2x)=0,0<= x<= 2pi ?

The general solution for sin(2x)=0,0<= x<= 2pi is x=0,x= pi/2 ,x=pi,x=(3pi)/2 ,x=2pi

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sin(2x)=0,0<= x<= 2pi

sin (2 x) = 0, 0 \leq x \leq 2 π

x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}, x = 2 π

Degrees

x = 0^{\circ}, x = 9 0^{\circ}, x = 18 0^{\circ}, x = 27 0^{\circ}, x = 36 0^{\circ}

Solution steps

sin (2 x) = 0, 0 \leq x \leq 2 π

General solutions for

sin (2 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}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = \frac{π}{2}, x = π, x = \frac{3 π}{2}, x = 2 π

\frac{1}{2} = sin^{2} (x)

- 12 sin^{2} (θ) + 3 = 0

2 cot^{2} (x) - 4 = 2

4 sin^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

4 sin (B) + 8 = 0

What is the general solution for sin(2x)=0,0<= x<= 2pi ?
The general solution for sin(2x)=0,0<= x<= 2pi is x=0,x= pi/2 ,x=pi,x=(3pi)/2 ,x=2pi