i need the answer and solution preferably please.
if tan(120∘−x)=sin(120∘)−sin(x)cos(120∘)−cos(x), where 0∘<x<180∘, compute x
Let s=sin(120∘)=sin(60∘)=√32Let c=cos(120∘)=−cos(60∘)=−12
LHS:
tan(120∘−x)=sin(120∘−x)cos(120∘−x)=sin(120∘)cos(x)−cos(120∘)sin(x)cos(120∘)cos(x)+sin(120∘)sin(x)=s⋅cos(x)−c⋅sin(x)c⋅cos(x)+s⋅sin(x)
RHS:
sin(120∘)−sin(x)cos(120∘)−cos(x)=2cos(120∘+x2)sin(120∘−x2)−2sin(120∘+x2)sin(120∘−x2)=−cos(120∘+x2)sin(120∘+x2)=−cos(60∘+x2)sin(60∘+x2)=−cos(60∘)cos(x2)−sin(60∘)sin(x2)sin(60∘)cos(x2)+cos(60∘)sin(x2)=sin(60∘)sin(x2)−cos(60∘)cos(x2)sin(60∘)cos(x2)+cos(60∘)sin(x2)=s⋅sin(x2)+c⋅cos(x2)s⋅cos(x2)−c⋅sin(x2)
s⋅cos(x)−c⋅sin(x)c⋅cos(x)+s⋅sin(x)=s⋅sin(x2)+c⋅cos(x2)s⋅cos(x2)−c⋅sin(x2)(s⋅cos(x)−c⋅sin(x))(s⋅cos(x2)−c⋅sin(x2))=(c⋅cos(x)+s⋅sin(x))(s⋅sin(x2)+c⋅cos(x2))s2cos(x)cos(x2)−sccos(x)sin(x2)−scsin(x)cos(x2)+c2sin(x)sin(x2)=c2cos(x)cos(x2)+sccos(x)sin(x2)+scsin(x)cos(x2)+s2sin(x)sin(x2)cos(x)cos(x2)(s2−c2)−sin(x)sin(x2)(s2−c2)−2sc⋅cos(x)sin(x2)−2sc⋅sin(x)cos(x2)=0(s2−c2)(cos(x)cos(x2)−sin(x)sin(x2)⏟=cos(x+x2))−2sc⋅(cos(x)sin(x2)+sin(x)cos(x2)⏟=sin(x+x2))=0(s2−c2)cos(x+x2)−2sc⋅sin(x+x2)=02sc⋅sin(x+x2)=(s2−c2)cos(x+x2)2sc⋅sin(3x2)=(s2−c2)cos(3x2)tan(3x2)=s2−c22sc|s2−c2=34−14=12tan(3x2)=122sc|2sc=2√32(−12)=−√32tan(3x2)=12−√32tan(3x2)=−√333x2=arctan(−√33)+n⋅180∘n∈Zx=23⋅arctan(−√33)+23⋅n⋅180∘x=23⋅(−30∘)+n⋅120∘x=−20∘+n⋅120∘n=1(0∘<x<180∘)x=−20∘+120∘x=100∘