solving an equation involving trig

√3 tan x + 2 sin x = 0   rewrite as

√3sin x / cos x + 2 sin x = 0  factor

sin x (√3 / cos x + 2 ) = 0       

And sin x  = 0   at  -180, 0 and 180 on the given interval

For he second factor, we have 

(√3 / cos x + 2 )  = 0       rewrite as  ( √3sec x + 2 ) = 0

So we have

sec x = -2 / √3   and this would be where the cos x = -√3 / 2

And this would occur at -150 and 150 on the given interval

Here's a graph of the solutions....

https://www.desmos.com/calculator/saykyobjdx