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

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

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sin(θ)=-0

sin (θ) = - 0

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = - 0

sin (θ) = 0

General solutions for

sin (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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

tan (x) = \frac{15}{12}

6 sin (θ) + 3 sin (θ) tan (θ) = 3 sin (θ)

cos (2 x) sin (x) = 1

sin (θ) = 10

17. 6^{2} = 1 5^{2} + 13. 1^{2} - 2 (15) (13.1) \cdot cos (A)

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