What is the general solution for 2sin(2θ)=0 ?

The general solution for 2sin(2θ)=0 is θ=pin,θ= pi/2+pin

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2sin(2θ)=0

2 sin (2 θ) = 0

θ = πn, θ = \frac{π}{2} + πn

Solution steps

2 sin (2 θ) = 0

Divide both sides by

2

sin (2 θ) = 0

General solutions for

sin (2 θ) = 0

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

θ = πn, θ = \frac{π}{2} + πn

- sin (2 x) = - 2 sin (x)

sin (8 x + 2) = cos (6 x - 10)

cos (2 θ - \frac{π}{2}) = - 1, 0 \leq θ \leq 2 π

sin^{2} (2 x) = 2 sin^{2} (x)

2 cos (x) + 2 sin (x) = 0

What is the general solution for 2sin(2θ)=0 ?
The general solution for 2sin(2θ)=0 is θ=pin,θ= pi/2+pin