(2sinθ+√3)(2cosθ-√3)=0 Write answer in radians in terms of pi?

sin Ө = - √3 / 2 , cos Ө = √3 / 2
Ө = 4π/3 , 5π/3 , π/6 , 11π/6

Let t replace theta. Put [sin(t),cos(t)] = (s,c). Then (2s+rt3)(2c-rt3) = 0--> s = -(1/2)rt3 ...[1] or
c = (1/2)rt3...[2]. Let 0­­­­­ (</=) t < 2pi. If [1] holds, t =4pi/3 or 5pi/3. If [2] holds, t =pi/6 or 11pi/6.
The 4 possible answers are t = pi/6, 4pi/3, 5pi/3, 11pi/6.

It's already factored. All you have to do is remember that the product of 0 and a number is 0. That means that if any of the factors are equaled to 0, then you've got a solution

2 * sin(t) + sqrt(3) = 0
2 * sin(t) = -sqrt(3)
sin(t) = -sqrt(3)/2

2 * cos(t) - sqrt(3) = 0
2 * cos(t) = sqrt(3)
cos(t) = sqrt(3)/2

These are principal values of the unit circle, which you need to learn.

On the unit circle, x = cos(t) and y = sin(t)

http://aleigha2016.weebly.com/uploads/3/8/7/5/38753757/4860426_orig.png

4pi/3, 5pi/3, 11pi/6, pi/6