#1 + sin^3 (x) + cos^3 (x) = 3/2 sin (2x)#
#=>1 + (sin (x) + cos (x))(sin^2x+cos^2(x)-sin(x)cos(x)) = 3/2 sin (2x)#
#=>1 + (sin (x) + cos (x))(1-1/2xx2sin(x)cos(x)) = 3/2 sin (2x)#
#=>1 + (sin (x) + cos (x))(1-1/2((sin(x)+cos(x))^2-1)) = 3/2 ((sinx+cosx)^2-1)#
Putting #sinx+cosx=y#
#=>1 + y(1-1/2(y^2-1) = 3/2 (y^2-1)#
#=> (y+1)-y/2(y-1)(y+1)-3/2 (y^2-1)=0#
#=> 2(y+1)-y(y-1)(y+1)-3 (y^2-1)=0#
#=> (y+1)(2-y(y-1)-3 (y-1))=0#
#=> (y+1)(5-y^2-2y)=0#
So when #y+1=0#
#sinx+cosx=-1#
#=>1/sqrt2cosx+1/sqrt2sinx=-1/sqrt2#
#=>cos(pi/4)cosx+sin(pi/4)sinx=cos(pi-pi/4)#
#=>cos(x-pi/4)=cos((3pi)/4)#
So
#x-pi/4=2npipm(3pi)/4" where "n inZZ#
Hence
#x=(2n+1)pi" where "n inZZ#
And
#x=2npi-(3pi)/4+pi/4" where "n inZZ#
#=>x=(4n-1)pi/2" where "n inZZ#
Again when #(5-y^2-2y)=0#
#=>y^2+2y-5=0#
#=>y=(-2pmsqrt(2^2-4*1(-5)))/2#
#=>y=(-2pm2sqrt6)/2#
#=>y=(-1pmsqrt6)#
So proceeding in similar way we get
#cos(x-pi/4)=(-1pmsqrt6)/sqrt2# which is #>1or<-1# hence the solution is rejected.
