Find the exact value? #2sinxcosx+sinx-2cosx=1#

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Answer 1 · 2018-05-09T10:51:09

#rarrx=2npi+-(2pi)/3# OR #x=npi+(-1)^n(pi/2)# where #nrarrZ#

#rarr2sinx*cosx+sinx-2cosx=1#

#rarrsinx(2cosx+1)-2cosx-1=#

#rarrsinx(2cosx+1)-1(2cosx+1)=0#

#rarr(2cosx+1)(sinx-1)=0#

Either, #2cosx+1=0#

#rarrcosx=-1/2=-cos(pi/3)=cos(pi-(2pi)/3)=cos((2pi)/3)#

#rarrx=2npi+-(2pi)/3# where #nrarrZ#

OR, #sinx-1=0#

#rarrsinx=1=sin(pi/2)#

#rarrx=npi+(-1)^n(pi/2)# where #nrarrZ#