# Calculus

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2sin(2theta) + sqrt(3) = 0
interval (0,2pi)

How do I solve this? I thought about maybe subtracting the sqrt(3) from both sides, so that i would have
2sin(2theta) = -sqrt(3)

Then maybe dividing by two?

sin(2theta) = -sqrt(3)/2

Am I doing anything right here? It's been a long time since I've worked with Trig, so I'm not sure . . .

• Calculus -

You correctly figured out
sin(2theta) = -sqrt(3)/2

That means that 2theta must be either 4 pi/3 or 5 pi/3 degrees (both of which have that sine value), or a multiple of 2 pi degrees added to those angles.

theta can therefore be
2 pi/3 or 5 pi/6.
Since they want it for an interval (0,2 pi), you can add pi to those values for two more solutions: 5 pi/3 and 11 pi/6.

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