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




Question Number 1961 by alib last updated on 26/Oct/15
Solve...    1+ sin 2x = sin x + cos x
$${Solve}… \\ $$$$ \\ $$$$\mathrm{1}+\:{sin}\:\mathrm{2}{x}\:=\:{sin}\:{x}\:+\:{cos}\:{x} \\ $$
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.
$$\mathrm{1}+\:{sin}\:\mathrm{2}{x}\:=\:{sin}\:{x}\:+\:{cos}\:{x} \\ $$$$\mathrm{1}+\mathrm{2}{sin}\:{x}\:{cos}\:{x}−{sin}\:{x}\:−{cos}\:{x}=\mathrm{0} \\ $$$${sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}+\mathrm{2}{sin}\:{x}\:{cos}\:{x}\:−\left({sin}\:{x}+{cos}\:{x}\right)=\mathrm{0}\: \\ $$$$\left({sin}\:{x}\:+{cos}\:{x}\right)^{\mathrm{2}} −\left({sin}\:{x}+{cos}\:{x}\right)=\mathrm{0} \\ $$$$\left({sin}\:{x}+{cos}\:{x}\right)\left({sin}\:{x}+{cos}\:{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${sin}\:{x}\:+\:{cos}\:{x}\:=\:\mathrm{0}\:\mid\:{sin}\:{x}+{cos}\:{x}−\mathrm{1}=\mathrm{0} \\ $$$${sin}\:{x}\:+\:{cos}\:{x}\:=\:\mathrm{0}\:\mid\:{sin}\:{x}\:+{cos}\:{x}=\mathrm{1} \\ $$$${sin}\:{x}=−{cos}\:{x}\:\:\:\:\:\:\:\mid\:{sin}\:{x}=\mathrm{1}−{cos}\:{x} \\ $$$${sin}^{\mathrm{2}} {x}={cos}^{\mathrm{2}} {x}\:\:\:\:\:\:\:\:\:\mid{sin}^{\mathrm{2}} {x}=\mathrm{1}+{cos}^{\mathrm{2}} {x}−\mathrm{2}\:{cos}\:{x} \\ $$$${sin}^{\mathrm{2}} {x}=\mathrm{1}−{sin}^{\mathrm{2}} {x}\:\:\:\mid\:\mathrm{1}−{cos}^{\mathrm{2}} {x}=\mathrm{1}+{cos}^{\mathrm{2}} {x}−\mathrm{2}\:{cos}\:{x} \\ $$$${sin}^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{cos}^{\mathrm{2}} {x}−{cos}\:{x}=\mathrm{0} \\ $$$${x}={sin}^{−\mathrm{1}} \left(\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:\:\:\:\:\:\mid\:{cos}\:{x}\left({cos}\:{x}−\mathrm{1}\right)=\mathrm{0}\Rightarrow{cos}\:{x}=\mathrm{0}\:\vee\:{cos}\:{x}=\mathrm{1} \\ $$$${x}=\frac{\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}},\frac{\mathrm{7}\pi}{\mathrm{4}}\:{in}\:{the}\:{interval}\:\left(\mathrm{0},\mathrm{2}\pi\right)\:\mid\:{x}=\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}},\mathrm{0}\:{in}\:\left(\mathrm{0},\mathrm{2}\pi\right) \\ $$$${Extraneous}\:{roots}\:{may}\:{be}\:{inluded}. \\ $$

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