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Question

Solve: 2(sinx-cos2x)-sin2x(1+2sinx)+2cosx=0


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Solution

Solving 2(sinx-cos2x)-sin2x(1+2sinx)+2cosx=0:

2sinxsin2x2cos2x2sinxsin2x+2cosx=0

We know that, sin2x=2sinxcosx

So, 2sinxsin2x=2sinx(2sinxcosx)

=4sin2xcosx

=4(1cos2x)cosx (sin2x=1-cos2x)

=4cosx4cos3x

=cosx+(3cosx4cos3x)

=cosx-cos3x (cos3x=-3cosx+4cos3x)

So, 2sinx2sinxcosx2cos2x(cosxcos3x)+2cosx=0

2sinx(1cosx)+4cos3x3cosx+cosx2(2cos2x1)=0

2sinx(1cosx)+4cos3x4cos2x2cosx+2=0

2sinx(1cosx)4cos2x(1cosx)+2(1cosx)=0

(1cosx)[2sinx4(1sin2x)+2]=0

cosx=1orsinx2(1sin2x)+1=0

x=2nπor2sin2x+sinx-1=0

2sin2x+sinx-1=0

2sin2x+2sinx-1sinx-1=0

2sinx(sinx+1)-(sinx+1)=0

(2sinx-1)(sinx+1)=0

x=mπ+(-1)mπ6orx=+(-1)k-π2

Hence general solutions are x=2nπorx=mπ+(-1)mπ6orx=+(-1)k-π2


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