#Sin2xcos2x-cos2x+sin2x-1=0# Solve for x?

Question

Find x such that 0 < x < 360

3952 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-20T12:17:55

#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#

Answer 2 · 2017-11-20T12:20:44

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#

Answer 3 · 2017-11-20T12:28:53

#x=45°, 90°,225°,270°#

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° #.