Solving Trigonometric Equations

Mar 2019
14
0
United States
Can anyone solve this and walk me through how you got your answer?

2sin(2x)-2sinx+2[FONT=DDG_ProximaNova]√3 cosx-[/FONT][FONT=DDG_ProximaNova]√3 =0[/FONT]
 

Plato

MHF Helper
Aug 2006
22,469
8,639
Can anyone solve this and walk me through how you got your answer?
2sin(2x)-2sinx+2√3 cosx-√3 =0
\(\displaystyle \left\{ \begin{array}{l}2\sin(2x)-2\sin(x)+2\sqrt3\cos(x)-\sqrt3=0\\4\sin(x)\cos(x)-2\sin(x)+2\sqrt3\cos(x)-\sqrt3=0\\2\sin(x)[2\cos(x)-1]-\sqrt3\left[2\cos(x)-1\right]=0\\\large(2\sin(x)-\sqrt3)(2\cos(x)-1)=0\end{array} \right.\)