What is the general solution for sqrt(3)sin(θ/2)+sin(θ)=0 ?

The general solution for sqrt(3)sin(θ/2)+sin(θ)=0 is θ=4pin,θ=2pi+4pin,θ=(5pi)/3+4pin,θ=(7pi)/3+4pin

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sqrt(3)sin(θ/2)+sin(θ)=0

3 sin (\frac{θ}{2}) + sin (θ) = 0

θ = 4 πn, θ = 2 π + 4 πn, θ = \frac{5 π}{3} + 4 πn, θ = \frac{7 π}{3} + 4 πn

Solution steps

3 sin (\frac{θ}{2}) + sin (θ) = 0

Let:

u = \frac{θ}{2}

3 sin (u) + sin (2 u) = 0

Rewrite using trig identities

sin (u) 3 + 2 cos (u) sin (u) = 0

Factor

sin (u) 3 + 2 cos (u) sin (u) : sin (u) (3 + 2 cos (u))

sin (u) (3 + 2 cos (u)) = 0

Solving each part separately

sin (u) = 0 or 3 + 2 cos (u) = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

3 + 2 cos (u) = 0 : u = \frac{5 π}{6} + 2 πn, u = \frac{7 π}{6} + 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn, u = \frac{5 π}{6} + 2 πn, u = \frac{7 π}{6} + 2 πn

Substitute back

u = \frac{θ}{2}

\frac{θ}{2} = 2 πn : θ = 4 πn

\frac{θ}{2} = π + 2 πn : θ = 2 π + 4 πn

\frac{θ}{2} = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{3} + 4 πn

\frac{θ}{2} = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{3} + 4 πn

θ = 4 πn, θ = 2 π + 4 πn, θ = \frac{5 π}{3} + 4 πn, θ = \frac{7 π}{3} + 4 πn

2 sin (x) cos (x) sin (x) = cos (x)

- 6 csc^{2} (x) + 10 cot (x) = - 10

x, cos (x) - cos (y) = 0

- 3 tan^{2} (θ) + 2 tan (θ) = 6 tan (θ) + 1

sin (x) + cos (x) = - 3

What is the general solution for sqrt(3)sin(θ/2)+sin(θ)=0 ?
The general solution for sqrt(3)sin(θ/2)+sin(θ)=0 is θ=4pin,θ=2pi+4pin,θ=(5pi)/3+4pin,θ=(7pi)/3+4pin