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)/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 < 2 π

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

Solution steps

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

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

Combine all the solutions

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

7 sin (x) = cos (x) - 4

5 sin^{2} (θ) - 11 sin (θ) + 2 = 0

tan^{2} (θ) + 3 sec (θ) = - 3

cos^{2} (x) - sin^{2} (x) = 1 - sin^{2} (x)

sin (2 x) cos (x) + sin (x) = 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)/3 ,θ=(4pi)/3