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Please solve it I attached it please provide answer as soon as possible

Pramod , 6 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 6 Years ago

Dear Pramod
2(sinx-cos2x)-2sinxcosx-2sin^2xcosx+2cosx=0
2(sinx-cos2x)-sin2x(1+2sinx)+2cosx=0
sinx-cos2x-sinxcosx-sin^2xcosx+cosx=0
sinx-(1-2sin^2x)-sinxcosx-sin^2xcosx+cosx=0   {cos2x=1-2sin^2x 
sinx-1+2sin^2x-sinxcosx-sin^2xcosx+cosx=0                and sin2x=2sinxcosx}
sinx-sinxcosx+2sin^2x-sin^2xcosx+cosx-1=0
sinx(1-cosx)+2sin^2x(1-cosx)-(1-cosx)=0
(1-cosx)[2sin^2x+sinx-1]=0
1-cosx=0 or 2sin^2x+sinx-1=0
now   2sin^2x+sinx-1=0 
     sinx=[-1+- root{1-4*2(-1)}]/2*2
   sinx=-1 and ½ 
now write general solution for cosx=1 ,sinx=-1 and sinx=1/2

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