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}

Solution steps

sin (2 x) = 0, - π \leq x \leq π

General solutions for

sin (2 x) = 0

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

Solve

2 x = 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