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# 2(sinx-cosx)-sin2x(1+2sinx)+2cosx=0 please find general solution of this trigonometric equation

Grade:11

## 1 Answers

Arun
25763 Points
2 years ago

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

hence x = 2n*pi
and
x = n*pi + (-1)n * pi/2
and
x = n * pi + (-1)n * pi /6

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