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

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

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-sin(2θ)=0

- sin (2 θ) = 0

θ = πn, θ = \frac{π}{2} + πn

Solution steps

- sin (2 θ) = 0

Divide both sides by

- 1

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

∣ cos (2 x) ∣ = 1

sin^{3} (x) = sin^{3} (x) tan (x)

- 4 sin^{2} (θ) - 7 sin (θ) + 7 = 0

2 cos^{2} (4 x) = 1

sin (2 x) cos (x) + cos (2 x) sin (x) = \frac{1}{2}

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