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

Degrees

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

Solution steps

- sin (2 θ) = 0

Divide both sides by

- 1

- sin (2 θ) = 0

Divide both sides by

- 1

\frac{- sin ( 2 θ )}{- 1} = \frac{0}{- 1}

Simplify

sin (2 θ) = 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) ∣ = 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