$\displaystyle 2\sin{2x}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3=0$
Solve for x.
Maple gave 60, 120, 150 & -150 degrees as the answers.
I tried to find a solution but in vain.
Hello, Kaloda!
$\displaystyle \text{Solve for }x\!:\;2\sin{2x}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3\:=\:0$
We have: .$\displaystyle 2\underbrace{\sin{2x}}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3\:=\:0$
Then: .$\displaystyle 2\overbrace{(2\sin x\cos x)} + 2\sqrt{3}\sin x - 2\sqrt{3}\cos x - 3 \:=\:0$
. . . . . . $\displaystyle 4\sin x\cos x + 2\sqrt{3}\sin x - 2\sqrt{3}\cos x - 3 \:=\:0$
Factor: .$\displaystyle 2\sin x\left(2\cos x + \sqrt{3}\right) - \sqrt{3}\left(2\cos x + \sqrt{3}\right) \:=\:0$
Factor: .$\displaystyle \left(2\cos x + \sqrt{3}\right)\left(2\sin x - \sqrt{3}\right) \:=\:0 $
Therefore:
. . $\displaystyle 2\cos x + \sqrt{3} \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\frac{\sqrt{3}}{2} \quad\Rightarrow\quad x \:=\:150^o,\,210^o$
. . $\displaystyle 2\sin x - \sqrt{3} \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\frac{\sqrt{3}}{2} \quad\Rightarrow\quad x \:=\:60^o,\,120^o$