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sinx-cosx-1-x-




Question Number 205218 by hardmath last updated on 13/Mar/24
∣ sinx∣ + ∣ cosx ∣ = 1  ⇒ x = ?
$$\mid\:\mathrm{sinx}\mid\:+\:\mid\:\mathrm{cosx}\:\mid\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:? \\ $$
Answered by Sutrisno last updated on 13/Mar/24
  (∣ sinx∣ + ∣ cosx ∣ )^2 = 1^2   sin^2 x+2∣sinx∣∣cosx∣+cos^2 x=1  ∣sin2x∣+1=1  ∣sin2x∣=0  2x=0+k.2π→x=k.π  2x=π−0+k.2π→x=(π/2)+k.π
$$ \\ $$$$\left(\mid\:\mathrm{sinx}\mid\:+\:\mid\:\mathrm{cosx}\:\mid\:\right)^{\mathrm{2}} =\:\mathrm{1}^{\mathrm{2}} \\ $$$${sin}^{\mathrm{2}} {x}+\mathrm{2}\mid{sinx}\mid\mid{cosx}\mid+{cos}^{\mathrm{2}} {x}=\mathrm{1} \\ $$$$\mid{sin}\mathrm{2}{x}\mid+\mathrm{1}=\mathrm{1} \\ $$$$\mid{sin}\mathrm{2}{x}\mid=\mathrm{0} \\ $$$$\mathrm{2}{x}=\mathrm{0}+{k}.\mathrm{2}\pi\rightarrow{x}={k}.\pi \\ $$$$\mathrm{2}{x}=\pi−\mathrm{0}+{k}.\mathrm{2}\pi\rightarrow{x}=\frac{\pi}{\mathrm{2}}+{k}.\pi \\ $$$$ \\ $$

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