Let S = [-2,2] : 2^2+3=0.

Question

Let $S = \{	heta\in[-2\pi,2\pi] : 2\cos^2	heta+3\sin	heta=0\}$. Then the sum of the elements of $S$ is:

Accepted Answer

$$2(1-\sin^2\theta)+3\sin\theta=0 \implies 2\sin^2\theta-3\sin\theta-2=0 \implies (2\sin\theta+1)(\sin\theta-2)=0 \implies \sin\theta=-\frac12$$

Within $[-2\pi,2\pi]$: $\theta = \pi+\tfrac\pi6,\ 2\pi-\tfrac\pi6,\ -\tfrac\pi6,\ -\pi+\tfrac\pi6$ (i.e. $\tfrac{7\pi}6,\tfrac{11\pi}6,-\tfrac\pi6,-\tfrac{5\pi}6$).

$$\text{Sum} = \frac{7\pi}6+\frac{11\pi}6-\frac\pi6-\frac{5\pi}6 = \frac{12\pi}6 = 2\pi$$

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