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

The general solution for sin(2θ)+sin(θ)=0,0<= θ<= 2pi is θ=0,θ=pi,θ=2pi,θ=(2pi)/3 ,θ=(4pi)/3

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sin(2θ)+sin(θ)=0,0<= θ<= 2pi

sin (2 θ) + sin (θ) = 0, 0 \leq θ \leq 2 π

θ = 0, θ = π, θ = 2 π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Solution steps

sin (2 θ) + sin (θ) = 0, 0 \leq θ \leq 2 π

Rewrite using trig identities

sin (θ) + 2 cos (θ) sin (θ) = 0

Factor

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

sin (θ) (2 cos (θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or 2 cos (θ) + 1 = 0

sin (θ) = 0, 0 \leq θ \leq 2 π : θ = 0, θ = π, θ = 2 π

2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π : θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Combine all the solutions

θ = 0, θ = π, θ = 2 π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

1 + cos (x) = 2 - cos (x)

cos^{2} (x) - 3 cos (x) - 4 = 0

2 sin^{2} (x) = sin (x) + 3

2 tan (θ) + 2 = 0

8 cos (θ) + 43 = 0

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