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

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = 9 0^{\circ} + 18 0^{\circ} n

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

Divide both sides by

2

2 θ = 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2}

Simplify

θ = πn

θ = πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{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