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

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

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

sin (2 x) = 0, - π \leq x \leq π

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

Degrees

x = 0^{\circ}, x = 9 0^{\circ}, x = 18 0^{\circ}, x = - 18 0^{\circ}, x = - 9 0^{\circ}

Solution steps

sin (2 x) = 0, - π \leq x \leq 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

- π \leq x \leq π

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

cos (2 x) - sin (2 x) = 1

2 cos (θ) = 3 cos (3 θ)

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

cos (θ) = 0.7071

2 cos (x) - 5 sin (x) = 0

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