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Can you help me with this problem: #sinx+cosx=sinxcosx#? Please?

2 Answers
Mar 17, 2017

Answer:

#x=pi+1/2arcsin(2-2sqrt2)#

Explanation:

#sinx+cosx=sinxcosx#

#(sinx+cosx)^2=sin^2xcos^2x#

#sin^2x+2sinxcosx+cos^2x=sin^2xcos^2x#

#1+2sinxcosx=sin^2xcos^2x#

Let #y=sinxcosx#

#1+2y=y^2#

#y^2-2y-1=0#

#y=1+-sqrt2#

#sinxcosx=1+-sqrt2#

#sin2x = 2sinxcosx therefore 1/2sin2x=sinxcosx#

#1/2sin2x=1+-sqrt2#

#sin2x=2+-2sqrt2#

#2+2sqrt2>1 therefore# it has no solutions

#sin2x=2-2sqrt2#

#2x=arcsin(2-sqrt2)#

#x=1/2arcsin(2-sqrt2)#

Checking this value, we see that the two sides of the equation don't have the same value. This is because #sinx <0# but #abs(cosx)>abs(sinx)#, so one side of the eqn is positive whilst the other is negative.

So we add #pi# to the value to make #sin# positive and #cos# negative but we also keep #abs(cosx)>abs(sinx)#, so now both sides are negative and, more importantly, have the same value.

So #x=pi+1/2arcsin(2-2sqrt2)#

Apr 17, 2017

Answer:

#x = pi+1/2 arcsin(2(1-sqrt(2)))#

Explanation:

#(sinx+cosx)^2=sin^2x+cos^2x+2sinx cosx= 1+2sinx cosx#

so

#(sinx cosx)^2-2sinxcosx-1=0#

so

#sinx cosx = (2 pm sqrt(4+4))/2#

so

#2sinxcosx=2(1-sqrt(2))#

but

#sin(2x)=2sinx cosx = 2(1-sqrt(2))#

then

#x = pi+1/2 arcsin(2(1-sqrt(2)))#