How do you find the exact value of $2 sin 2 θ + sin θ = 0$ $2sin2theta+sintheta=0$ in the interval $0 \leq θ < 2 π$ $0<=theta<2pi$ ?

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How do you find the exact value of $2 sin 2 θ + sin θ = 0$ $2sin2theta+sintheta=0$ in the interval $0 \leq θ < 2 π$ $0<=theta<2pi$ ?

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Answer 1 · 2017-06-16T07:17:33

$θ = {0, 1.82, 3.14, 4.46}$ $theta={0,1.82,3.14,4.46}$ (in radians up to $2 d p$ $2dp$ )

Explanation:

As $2 sin 2 θ + sin θ = 0$ $2sin2theta+sintheta=0$ , we have

$2 \times 2 sin θ cos θ + sin θ = 0$ $2xx2sinthetacostheta+sintheta=0$

or $sin θ (4 cos θ + 1) = 0$ $sintheta(4costheta+1)=0$

hence either $sin θ = 0$ $sintheta=0$ i.e. $θ = 0$ $theta=0$ or $π$ $pi$ (in radians)

or $4 cos θ + 1 = 0$ $4costheta+1=0$ i.e. $cos θ = - \frac{1}{4} = cos 1.82$ $costheta=-1/4=cos1.82$ or $cos (2 π - 1.82) = cos 4.46$ $cos(2pi-1.82)=cos4.46$ (in radians up to $2 d p$ $2dp$ )

i.e. $θ = {0, 1.82, 3.14, 4.46}$ $theta={0,1.82,3.14,4.46}$

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How do you find the exact value of $2 sin 2 θ + sin θ = 0$ $2sin2theta+sintheta=0$ in the interval $0 \leq θ < 2 π$ $0<=theta<2pi$ ?

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Explanation:

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