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

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

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2sin(θ)+sin(2θ)=0

2 sin (θ) + sin (2 θ) = 0

θ = 2 πn, θ = π + 2 πn

Solution steps

2 sin (θ) + sin (2 θ) = 0

Rewrite using trig identities

2 sin (θ) + 2 cos (θ) sin (θ) = 0

Factor

2 sin (θ) + 2 cos (θ) sin (θ) : 2 sin (θ) (cos (θ) + 1)

2 sin (θ) (cos (θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or cos (θ) + 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

cos (θ) + 1 = 0 : θ = π + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

\frac{π}{4} = arctan (x)

2 cos (x) = cos (2 x)

3 cos^{2} (θ) + cos (θ) = 0

cos^{2} (θ) = 0.75

cos (y_{0}) = \frac{( a _{0} \cdot n _{0} )}{∣} a_{0} ∣

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