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

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

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asin(2θ)=0

a sin (2 θ) = 0

θ = πn, θ = \frac{π}{2} + πn

Solution steps

a sin (2 θ) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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) = 0, x \leq 2 π, 0

7 + csc (2 x) = 3 cot^{2} (2 x)

cos (θ) = cos (\frac{π}{13})

sin (2 x + \frac{π}{6}) = - 1

tan (x) = 2 π

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