#Sin2xcos2x-cos2x+sin2x-1=0#
Solve for x?
Find x such that 0 < x < 360
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#x_1=45#, #x_2=90#, #x_3=225#, #x_4=270#
#sin2x*cos2x-cos2x+sin2x-1=0#
#cos2x*(sin2x-1)+sin2x-1=0#
#(cos2x+1)*(sin2x-1)=0#
After equating multpliers to 0, I found
For #cos2x=-1#, or #2x=180+360k#, so #x_1=90# and #x_2=270#
For #sin2x=1#, or #2x=90+360k#, so #x_3=45# and #x_4=225#
There are four solutions in the interval #x:[0^o, 360^o]#
#x_1 = 90^o; \qquad x_2 = 45^o; \qquad x_3=270^o; \qquad x_4=225^o;#
#\sin2x\cos2x-\cos2x+\sin2x-1=0#
#\cos2x(\sin2x-1)+(\sin2x-1)=0#
#(\cos2x+1)(\sin2x-1)=0#
Either #(\cos2x+1)# vanishes or #(\sin2x-1)# vanishes...
Solutions for #(\cos2x+1)=0# in the interval #[0^o, 360^o]# are #x_1=90^o; \qquad x_2=270^o#
Solutions for #(\sin2x-1)=0# in the interval #[0^o, 360^o]# are #x_3=45^o; \qquad x_4=225^o#
Here, #sin2x*cos2x-cos2x+sin2x-1=0#
or, #cos2x(sin2x-1)+1(sin2x-1)=0#
or, #(cos2x+1)(sin2x-1)=0#
Either #cos2x+1=0#......(1)
OR, #sin2x-1=0#.......(2)
From (1) we have
#cos2x=-1=cos180°=cos540°#
or, #x=90°,270°#
From (2) we have
#sin2x-1=0#
or, #sin2x=1=sin90°=sin450°#
or, #2x=90°, 450°#
or, #x=45°, 225°#
So, the values of #x# are #45° ,90°, 225° and 270° #.
