Solve: 1-sin2x = cosx- sinx

We have to solve 1−sin2x=cosx−sinx

Solution

1−sin2x=cosx−sin x

(cosx−sin x) 2=cos x−sin x

So, cos x−sinx=1

tanx=1,

x=nπ+ π/4

cos x−sinx=1
and cos x−1=sinx

−2sin 2 (x/2) =2sin(x/2) cos (x/2)

sin (x/2)= 0 and x/2=n∏

x= 2n∏

so tan (x/2)=-1

x/2 = n∏ – ∏/4

x= 2n∏ – ∏/2

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