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

Solution steps

sin (2 x) = 0, 0 \leq x \leq 2 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

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