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Solve-1-sin-2x-sin-x-cos-x-

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Question Number 1961 by alib last updated on 26/Oct/15
Solve... 1+ sin 2x = sin x + cos x
Solve…1+sin2x=sinx+cosx
Answered by Rasheed Soomro last updated on 26/Oct/15
1+ sin 2x = sin x + cos x 1+2sin x cos x−sin x −cos x=0 sin^2 x+cos^2 x+2sin x cos x −(sin x+cos x)=0 (sin x +cos x)^2 −(sin x+cos x)=0 (sin x+cos x)(sin x+cos x−1)=0 sin x + cos x = 0 ∣ sin x+cos x−1=0 sin x + cos x = 0 ∣ sin x +cos x=1 sin x=−cos x ∣ sin x=1−cos x sin^2 x=cos^2 x ∣sin^2 x=1+cos^2 x−2 cos x sin^2 x=1−sin^2 x ∣ 1−cos^2 x=1+cos^2 x−2 cos x sin^2 x=(1/2) ∣ cos^2 x−cos x=0 x=sin^(−1) (±(1/( (√2)))) ∣ cos x(cos x−1)=0⇒cos x=0 ∨ cos x=1 x=(π/4),((3π)/4),((5π)/4),((7π)/4) in the interval (0,2π) ∣ x=(π/2),((3π)/2),0 in (0,2π) Extraneous roots may be inluded.
1+sin2x=sinx+cosx1+2sinxcosx−sinx−cosx=0sin2x+cos2x+2sinxcosx−(sinx+cosx)=0(sinx+cosx)2−(sinx+cosx)=0(sinx+cosx)(sinx+cosx−1)=0sinx+cosx=0∣sinx+cosx−1=0sinx+cosx=0∣sinx+cosx=1sinx=−cosx∣sinx=1−cosxsin2x=cos2x∣sin2x=1+cos2x−2cosxsin2x=1−sin2x∣1−cos2x=1+cos2x−2cosxsin2x=12∣cos2x−cosx=0x=sin−1(±12)∣cosx(cosx−1)=0⇒cosx=0∨cosx=1x=π4,3π4,5π4,7π4intheinterval(0,2π)∣x=π2,3π2,0in(0,2π)Extraneousrootsmaybeinluded.

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