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

The general solution for sin(3x)=0,0<= x<2pi is x=0,x= pi/3 ,x=(2pi)/3 ,x=pi,x=(4pi)/3 ,x=(5pi)/3

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sin(3x)=0,0<= x<2pi

sin (3 x) = 0, 0 \leq x < 2 π

x = 0, x = \frac{π}{3}, x = \frac{2 π}{3}, x = π, x = \frac{4 π}{3}, x = \frac{5 π}{3}

Degrees

x = 0^{\circ}, x = 6 0^{\circ}, x = 12 0^{\circ}, x = 18 0^{\circ}, x = 24 0^{\circ}, x = 30 0^{\circ}

Solution steps

sin (3 x) = 0, 0 \leq x < 2 π

General solutions for

sin (3 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}

3 x = 0 + 2 πn, 3 x = π + 2 πn

3 x = 0 + 2 πn, 3 x = π + 2 πn

Solve

3 x = 0 + 2 πn : x = \frac{2 πn}{3}

3 x = 0 + 2 πn

0 + 2 πn = 2 πn

3 x = 2 πn

Divide both sides by

3

3 x = 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{2 πn}{3}

Simplify

x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

Solve

3 x = π + 2 πn : x = \frac{π}{3} + \frac{2 πn}{3}

3 x = π + 2 πn

Divide both sides by

3

3 x = π + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{π}{3} + \frac{2 πn}{3}

Simplify

x = \frac{π}{3} + \frac{2 πn}{3}

x = \frac{π}{3} + \frac{2 πn}{3}

x = \frac{2 πn}{3}, x = \frac{π}{3} + \frac{2 πn}{3}

Solutions for the range

0 \leq x < 2 π

x = 0, x = \frac{π}{3}, x = \frac{2 π}{3}, x = π, x = \frac{4 π}{3}, x = \frac{5 π}{3}

2 cos^{2} (x) = 3 cos (x)

sin (θ) = - \frac{1}{5}, π < θ < \frac{3 π}{2}, cos (θ)

3 cot (θ) - 1 = 0, 0 \leq θ \leq 2 π

\frac{sin ( x )}{37} = \frac{sin ( 3 0 ^{\circ} )}{30}

1 - cos (x) = cos (x)

What is the general solution for sin(3x)=0,0<= x<2pi ?
The general solution for sin(3x)=0,0<= x<2pi is x=0,x= pi/3 ,x=(2pi)/3 ,x=pi,x=(4pi)/3 ,x=(5pi)/3