need help asap

Are you SURE 30 degrees solves this equation?  I don't think it does.....

2sin(x)tan(x)=tan(x)+2sin(x)

\(2sin(x)tan(x)=tan(x)+2sin(x)\\ \frac{2sin^2(x)}{cos(x)}=\frac{sin(x)}{cos(x)}+\frac{2sin(x)cos(x)}{cos(x)}\\\ 2sin^2(x)=sin(x)+2sin(x)cos(x)\\ 2sin(x)-2cos(x)=1\\ 4(sin(x)-cos(x))^2=1\\ 4(sin^2(x)-2sinxcosx+cos^2(x))=1\\ sin^2(x)+cos^2(x)-2sinxcosx=0.25\\ 1-2sinxcosx=0.25\\ -2sinxcosx=\;-0.75\\ 2sinxcosx=\;0.75\\ sin(2x)=\frac{3}{4}\\ 2x\mbox{ must be in the 1st ot 2nd quadrant}\\ 2x=asin(\frac{3}{4})\\ 2x\approx 48.6^\circ,\quad 180-48.6^\circ, \quad (48.6+360)^\circ,\quad (180-48.6+360)^\circ, \quad \\ 2x\approx 48.6^\circ,\quad 131.4^\circ, \quad 408.6^\circ,\quad 491.4^\circ, \quad \\ x\approx 24.3^\circ,\quad 65.7^\circ, \quad 204.3^\circ,\quad 245.7^\circ, \quad \\ \)

I have checked these points from substitution.  Only the 2 middle ones work. It has probably got something to do with the squaring that I did as part of the working

\(x= 65.7^\circ \qquad or \qquad 204.3^\circ \qquad (\pm360n)\)