How do you solve $sin (2 θ) + sin θ = 0$ $sin(2theta)+sintheta=0$ on the interval $0 \leq θ < 2 π$ $0<=theta<2pi$ ?

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Answer 1 · 2015-04-27T18:22:30

$θ = 0, \frac{2 π}{3}, π or \frac{4 π}{3}$ $theta=0, {2pi}/3, pi or {4pi}/3$

Use $sin (2 θ) = 2 sin θ cos θ$ $sin(2 theta) = 2 sintheta costheta$

Then factor and solve.

$sin (2 θ) + sin θ = 0$ $sin(2 theta) + sintheta =0$

$2 sin θ cos θ + sin θ = 0$ $2 sintheta costheta + sintheta =0$

$sin θ (2 cos θ + 1) = 0$ $sin theta (2cos theta +1) =0$

$sin θ = 0$ $sin theta = 0$ $sssss$ $color(white)"sssss"$ or $sssss$ $color(white)"sssss"$ $2 cos θ + 1 = 0$ $2 cos theta + 1 =0$ so

$sin θ = 0$ $sin theta = 0$ $sssss$ $color(white)"sssss"$ or $sssss$ $color(white)"sssss"$ $cos θ = - \frac{1}{2}$ $cos theta = -1/2$

$θ = 0, \frac{2 π}{3}, π or \frac{4 π}{3}$ $theta=0, {2pi}/3, pi or {4pi}/3$

How do you solve sin(2theta)+sintheta=0sin(2θ)+sinθ=0sin(2theta)+sintheta=0 on the interval 0<=theta<2pi0≤θ<2π0<=theta<2pi?